1. A speed boat is racing across a lake at 25 meters per
second when its motor burns out. It then slowly
comes to a stop over the next 45 seconds. What was its
acceleration?
v = u + a t
where u = initial velocity (25 m/s), v = final velocity (0), a = acceleration, and t = time (45 s). So
0 = 25 m/s + a (45 s)
a = (-25 m/s) / (45 s)
a ≈ -0.56 m/s²
The feeling of weightlessness occurs because _____________________.
there is no supporting force under your mass.
there is no gravity present.
there is only a small amount of gravity present.
Answer:
there is only a small amount of gravity present.
Explanation:
this is because the only force acting upon your body during free fall is the force of gravity which is a non contact force.
The Jamaican Bobsled Team is sliding down a hill in a toboggan at a rate of 5 m/s when he reaches an even steeper slope. If he accelerates at 2 m/s2 for the 5 m slope, how fast is he traveling when he reaches the bottom of the 5 m slope?
Answer:
6.7 m/s
Explanation:
Given:
Δx = 5 m
v₀ = 5 m/s
a = 2 m/s²
Find: v
v² = v₀² + 2aΔx
v² = (5 m/s)² + 2 (2 m/s²) (5 m)
v = 6.7 m/s
“Permafrost” is permanently frozen soil and occurs mostly in high latitudes storing a massive
amount of a particular element. As a result of climate change, permafrost is at the risk of melting and
releasing the stored element in the form of a gas. Identify the gas.
a) Ozone
b) Hydrogen
c) Nitrogen oxide
d) Carbon dioxide
A soccer ball accelerates from rest and rolls 6.5m down a hill in 3.1 s. It then bumps into a tree. What is the speed of the ball just before it hits the tree.
Answer:
2.096m/s
Explanation:
The speed of this soccer ball can be calculated using the formula;
Speed = distance/time
According to this question, the distance of the ball before it hits the tree is 6.5m, the time it takes is 3.1s, hence;
Speed = 6.5/3.1
Speed of the ball = 2.096m/s
Therefore, the speed of the ball before hitting the tree is 2.096m/s
If the loudness drops to 90 % of its original value in 5.0 s , what is the time constant of the damped oscillation
This question is incomplete, the complete question is;
A gong makes a loud noise when struck. The noise gradually gets less and less loud until it fades below the sensitivity of the human ear. The simplest model of how the gong produces the sound we hear treats the gong as a damped harmonic oscillator. The tone we hear is related to the frequency f of the oscillation, and its loudness is proportional to the energy of the oscillation.
If the loudness drops to 90 % of its original value in 5.0 s , what is the time constant of the damped oscillation
Answer: the time constant of the damped oscillation is 47.44s
Explanation:
Given that;
t = 5.0s
Lets say Ao is the amplitude of initial loudness and later A(t) = 0.9 Ao
EXPRESSION for amplitude is A(t) = Ao e^-t / T
t is time while T is time constant
so
0.9Ao = Ao e^-t / T
0.9 = e^ -t/T
So we take the natural log of both the sides
ln (0.9) = -t/T
-0.1054 = -t/T
0.1054 = t/T
WE now substitute our value of t
0.1054 = t/T
0.1054T = 5.0
T = 5 / 0.1054
T = 47.44s
therefore the time constant of the damped oscillation is 47.44s
A man of weight 60N climbed stairs of height 15m high in 15s. Find the power
of the man,
The particles of a more dense substance are closer together
than the particles of a less dense substance.
TRUE
FALSE
The particles of a more dense substance are closer together than the particles of a less dense substance. Thus, the given statement is true.
What is density of particles?Density of the particles is the substance's mass per unit of volume. The symbol which is most often used for the density is ρ (rho), although the Latin letter D can also be used to denote density.
Density is the mass of a unit volume of a material substance or particle. The formula for density is d = M/V (mass per unit volume), where d is density, M is the mass of particle, and V is the volume. Density is commonly expressed in the units of grams per cubic centimeters.
The S.I. unit of density is made up of the mass of the particle which is kg and that of volume is meter cube. Hence, the S.I. unit of density is kg/m³.
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Question 16 1 pts Jessie feels pressured by his parents to get a job. This is an example of the law of?
readiness
disuse
effect
belonging
What is the maximum torque on a 150-turn square loop of wire 18.0 cm on a side that carries a 50.9 A current in a 1.60 T field
Answer:
The maximum torque on the loop is 395.80 N.m.
Explanation:
Given;
number of turns of the wire, N = 150 turns
length of the square loop, L = 18.0 cm = 0.18 m
current in the wire, I = 50.9 A
Magnetic field, B = 1.6 T
Maximum torque on the loop is given by;
τ = NIAB
τ = (150)(50.9)(0.18²)(1.6)
τ = 395.80 N.m
Therefore, the maximum torque on the loop is 395.80 N.m.
Please help me! What is Ohm's law?
Ohm's law shows the relationship between voltage, current, and the resistance of a energy bond. The formula for Ohm's law is:
voltage = current x resistance
This formula tells you that current and resistance is the voltage of an energy bond.
Hope this helps you!
Answer: What is Ohm’s Law?
Ohm's Law is a formula used to calculate the relationship between voltage, current and resistance in an electrical circuit.
Ohm's Law (E = IR) is as fundamentally important as Einstein's Relativity equation (E = mc²) is to physicists.
E = I × R
It means voltage = current × resistance, or volts = amps × ohms, or V = A × Ω.
Resistance cannot be measured in an operating circuit, so Ohm's Law is especially useful when it needs to be calculated. Rather than shutting off the circuit to measure resistance, a technician can determine R using the above variation of Ohm's Law.
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What type of observation is made through interviewing people’s
Answer:
Interviewing and observation are two methods of collecting qualitative data as part of research. ... Interviews vary from structured, in which a set list of questions is asked of every interviewee, to unstructured, which is open-ended.
What is the moment of inertia I of an object that rotates at 13.0 rev/min13.0 rev/min about an axis and has a rotational kinetic energy of 16.0 J?
Answer:
The moment of inertia of the object is 17.276 kilogram-square meters.
Explanation:
According to the statement, we find that object has rotation and no translation. From Rotation Physics we get that rotational kinetic energy ([tex]K_{R}[/tex]), measured in joules, is represented by the following formula:
[tex]K_{R} = \frac{1}{2}\cdot I_{G}\cdot \omega^{2}[/tex] (Eq. 1)
Where:
[tex]I_{G}[/tex] - Moment of inertia with respect to center of mass, measured in kilogram-square meters.
[tex]\omega[/tex] - Angular speed, measured in radians per second.
Now we clear the moment of inertia:
[tex]I_{G} = \frac{2\cdot K_{R}}{\omega^{2}}[/tex]
If we know that [tex]K_{R} = 16\,J[/tex] and [tex]\omega \approx 1.361\,\frac{rad}{s}[/tex], then the moment of inertia of the object is:
[tex]I_{G} = \frac{2\cdot (16\,J)}{\left(1.361\,\frac{rad}{s} \right)^{2}}[/tex]
[tex]I_{G} =17.276\,kg\cdot m^{2}[/tex]
The moment of inertia of the object is 17.276 kilogram-square meters.
The moment of inertia of the object will be "17.276 kg/m²".
Moment of inertiaRotational Kinetic energy, [tex]K_R[/tex] = 16 J
Angular speed, ω = 1.361 rad/s
By using the Rotation Physics, the relation will be:
→ [tex]K_R[/tex] = [tex]\frac{1}{2}[/tex] × [tex]I_G[/tex] × ω²
the,
The moment of inertia be:
→ [tex]I_G[/tex] = [tex]\frac{2\times K_R}{\omega^2}[/tex]
By substituting the values, we get
= [tex]\frac{2\times 16}{(1.361)^2}[/tex]
= [tex]\frac{32}{(1.361)^2}[/tex]
= 17.276 kg.m²
Thus the above answer is correct.
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gold has a density of 19.32g/cm3. if you have a 25 cm3 sample of gold what is the mass of the sample
Answer:
ggggggggggggggggggggggggggggg
Explanation:
Answer:
The volume of the sample of gold is
16.51 [tex]cm^{3}[/tex]
Explanation:
The formula for density is:
D= [tex]\frac{M}{V}[/tex].
where:
D is density,
M is mass, and
V is volume.
Rearrange the density formula to isolate volume.
V= [tex]\frac{M}{D}[/tex]
V= [tex]\frac{318.97g Au}{19.32g cm^{3}}[/tex]
V= 318.97∅ × [tex]\frac{1 cm^{3} Au}{19.32g cm^{3} }[/tex]← Multiply by the multiplicative inverse of the density.
V= 16.51 cm³ Au.
One airplane is approaching an airport from the north at 181 kn/hr. A second airplane approaches from the east at 278 km/hr. Find the rate at which the distance between the planes changes when the southbound plane is 30 km away from the airport and the westbound plane is 15 km from airport.
Answer:
The value is [tex] \frac{dR}{dt} = -286.2 \ km/hr [/tex]
Explanation:
From the question we are told that
The speed of the airplane from the north is [tex]\frac{dN}{dt} = -181 \ km /hr[/tex]
The negative sign is because the direction is towards the south
The speed of the airplane from the east is [tex]\frac{dE}{dt} = -278 \ km/hr[/tex]
The negative sign is because the direction is towards the west
The distance of the southbound plane from the airport is [tex]N = 30 \ km[/tex]
The distance of the westbound plane is [tex]E = 15 \ km[/tex]
Generally the distance between the plane is mathematically represented using Pythagoras theorem as
[tex]R^2 = N^2 + E^2[/tex]
Next differentiate implicitly this equation to obtain the rate at which the distance between the planes changes
So
[tex]2R\frac{dR}{dt} = 2N \frac{dN}{dt} + 2E\frac{dE}{dt}[/tex]
Here
[tex]R = \sqrt{N^2 + E^2}[/tex]
=> [tex]R = \sqrt{30^2 + 15^2}[/tex]
=> [tex]R = \sqrt{30^2 + 15^2}[/tex]
=> [tex]R =33.54 \ m [/tex]
[tex]2(33.54) * \frac{dR}{dt} = 2( 30)*(-181) + 2*15*(-278)[/tex]
=> [tex] 67.08 * \frac{dR}{dt} = -19200[/tex]
=> [tex] \frac{dR}{dt} = -286.2 \ km/hr [/tex]
The rate of change of the distance between the planes is 286.23 km/hr.
The given parameters;
speed of the airplane from North, dn/dt = 181 Km/hspeed of the airplane from the East, de/dt = 278 km/hnorth distance, n = 30 kmeast distance, e= 15 kmThe distance between the two planes is calculated by applying Pythagoras theorem as shown below;
[tex]d^2 = n^2 + e^2\\\\d = \sqrt{n^2 + e^2} \\\\d = \sqrt{30^2 + 15^2} \\\\d = 33.54 \ km[/tex]
The rate of change of the distance between the planes is calculated as follows;
[tex]d^2 = e^2 + n^2\\\\2\frac{dd}{dt} = 2e\frac{de}{dt} + 2n\frac{dn}{dt} \\\\d\frac{dd}{dt} = e\frac{de}{dt} + n\frac{dn}{dt}\\\\(33.54) \frac{dd}{dt} = (15)(278) \ + (30)(181)\\\\(33.54) \frac{dd}{dt} = 9600\\\\\frac{dd}{dt} = \frac{9600}{33.54} \\\\\frac{dd}{dt} = 286.23 \ km/hr[/tex]
Thus, the rate of change of the distance between the planes is 286.23 km/hr.
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The mass of a dropped object impacts its final velocity but not its acceleration.
Explanation:
because of the acceleration due to gravity is constant which is 9.8
Answer:
TRUE
Explanation:
CUZ , IT DOSEN'T AFFECT
Mass does not affect the speed of falling objects, assuming there is only gravity acting on it. Both bullets will strike the ground at the same time.
Paragraph/Comprehension type questions.
A body weighs 500gf in air and 300gf when completely immersed in water
36. Find the apparent loss in weight of the body.
1)500gf
2)300gf
3)200gf
4)800gf
37. Find the buoyant force acting on the body
1)500gf
2)300gf
3)200gf
4)800gf
Answer:
1>500gf
1>300gf
its answer
waht is science
wjwissbsskdldmndndnd
Answer:
the intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment.
Explanation:
A .05 kg rubber ball is dropped and hits the floor with an initial velocity of 10 m/s. It rebounds away from the floor with a final speed of 7 m/s after being in contact with the floor for .01 seconds. Find the magnitude of the force exerted by the floor on the rubber ball.
Answer:the answer is 3
Explanation:
A sound wave can be considered as a displacement wave or a pressure wave? What phase difference exists between the displacement and the pressure of a sound wave?
The equation that governs the period of a pendulum’s swinging. T=2π√L/g
Where T is the period, L is the length of the pendulum and g is a constant, equal to 9.8 m/s2. The symbol g is a measure of the strength of Earth’s gravity, and has a different value on other planets and moons.
On our Moon, the strength of earth’s gravity is only 1/6th of the normal value. If a pendulum on Earth has a period of 4.9 seconds, what is the period of that same pendulum on the moon?
Answer:
The period of that same pendulum on the moon is 12.0 seconds.
Explanation:
To determine the period of that same pendulum on the moon,
First, we will determine the value of g (which is a measure of the strength of Earth's gravity) on the Moon. Let the value of g on the Moon be [tex]g_{M}[/tex].
From the question, the strength of earth’s gravity is only 1/6th of the normal value. The normal value of g is 9.8 m/s²
∴ [tex]g_{M}[/tex] = [tex]\frac{1}{6} \times 9.8 m/s^{2}[/tex]
[tex]g_{M}[/tex] = 1.63 m/s²
From the question, T=2π√L/g
[tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]
We can write that,
[tex]T_{E} = 2\pi \sqrt{\frac{L}{g_{E} } }[/tex] .......... (1)
Where [tex]T_{E}[/tex] is the period of the pendulum on Earth and [tex]g_{E}[/tex] is the measure of the strength of Earth's gravity
and
[tex]T_{M} = 2\pi \sqrt{\frac{L}{g_{M} } }[/tex] .......... (2)
Where [tex]T_{M}[/tex] is the period of the pendulum on Moon and [tex]g_{M}[/tex] is the measure of the strength of Earth's gravity on the Moon.
Since we are to determine the period of the same pendulum on the moon, then, [tex]2\pi[/tex] and [tex]L[/tex] are constants.
Dividing equation (1) by (2), we get
[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]
From the question,
[tex]T_{E} = 4.9secs[/tex]
[tex]g_{E}[/tex] = 9.8 m/s²
[tex]g_{M}[/tex] = 1.63 m/s²
[tex]T_{M}[/tex] = ??
From,
[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]
[tex]\frac{4.9}{T_{M} } = \sqrt{\frac{1.63}{9.8} }[/tex]
[tex]\frac{4.9}{T_{M} } = 0.40783[/tex]
[tex]T_{M} =\frac{4.9}{0.40783 }[/tex]
[tex]T_{M} = 12.01 secs[/tex]
∴ [tex]T_{M} = 12.0secs[/tex]
Hence, the period of that same pendulum on the moon is 12.0 seconds.
Answer:
The period of that same pendulum on the moon is 12.0 s
Explanation:
Given;
period of a pendulum’s swinging, T=2π√L/g
the strength of earth’s gravity on moon, g₂ = ¹/₆(g₁)
period of pendulum on Earth, T₁ = 4.9 s
period of pendulum on moon, T₂ = ?
The length of the pendulum is constant, make it the subject of the formula;
[tex]T = 2\pi \sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g}}\\\\(\frac{T}{2\pi} )^2 =\frac{L}{g}\\\\\frac{T^2}{4\pi^2} = \frac{L}{g}\\\\ L = \frac{gT^2}{4\pi^2}\\\\L_1 = L_2\\\\\frac{g_1T_1^2}{4\pi^2}= \frac{g_2T_2^2}{4\pi^2}\\\\g_1T_1^2 = g_2T_2^2\\\\T_2^2 = \frac{g_1T_1^2}{g_2} \\\\T_2 = \sqrt{\frac{g_1T_1^2}{g_2}}\\\\ T_2 = \sqrt{\frac{g_1T_1^2}{g_1/6}}\\\\ T_2 = \sqrt{\frac{6*g_1T_1^2}{g_1}}\\\\T_2 = \sqrt{6T_1^2}\\\\ T_2 = T_1\sqrt{6} \\\\T_2 = (4.9)\sqrt{6}\\\\ T_2 = 12.0 \ s[/tex]
Therefore, the period of that same pendulum on the moon is 12.0 s
Assume that the particle has initial speed viviv_i. Find its final kinetic energy KfKfK_f in terms of viviv_i, MMM, FFF, and DDD.
Answer:
KE= 1/2mv²
Explanation:
The kinetic energy of a body is the energy possessed by virtue of the body in motion
Given the parameters
m which is the mass of the body
v which is the velocity of the body too
K.E = kinetic energy
The expression for the kinetic energy of a body is given as
KE= 1/2mv²
A sleigh is being pulled horizontally by a train of horses at a constant speed of 6.38 m/s. The magnitude of the normal force exerted by the snow-covered ground on the sleigh is 7.50 ✕ 103 N.
(a) If the coefficient of kinetic friction between the sleigh and the ground is 0.26, what is the magnitude of the kinetic friction force experienced by the sleigh?
N
(b) If the only other horizontal force exerted on the sleigh is due to the horses pulling the sleigh, what must be the magnitude of this force?
N
Answer:
(a). The kinetic friction force is 1950 N.
(b). The magnitude of force will be equal of friction force
Explanation:
Given that,
Constant speed = 6.38 m/s
Force [tex]F=7.50\times10^{3}\ N[/tex]
Kinetic friction = 0.26
(a). We need to calculate the friction force
Using formula of friction force
[tex]f_{k}=\mu F_{N}[/tex]
Put the value into the formula
[tex]f_{k}=0.26\times7.50\times10^{3}[/tex]
[tex]f_{k}=1950\ N[/tex]
(b). If the only other horizontal force exerted on the sleigh is due to the horses pulling the sleigh,
We need to calculate the magnitude of this force
According to given data,
The same force will be applied to keep constant velocity.
Hence, (a). The kinetic friction force is 1950 N.
(b). The magnitude of force will be equal of friction force.
(a). The kinetic friction force is 1950 N.
(b). The magnitude of force will be equal of friction force
The calculation is as follows;a. The magnitude of the kinetic friction force experienced by the sleigh is
[tex]= 0.76 \times 7.50 \times 10^3[/tex]
= 1950 N
b. It should be equivalent to the friction force.
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A single-turn circular loop of radius 9.4 cm is to produce a field at its center that will just cancel the earth's field of magnitude 0.7 G directed at 70degrees below the horizontal north direction. Find the current in the loop.
Answer:
The current in the loop is 10.5 A.
Explanation:
Given that,
Radius = 9.4 cm
Magnetic field = 0.7 G
Angle = 70°
We know that,
The magnetic field due to the current in a loop is
[tex]B_{I}=\dfrac{\mu_{0}NI}{2r}[/tex]
The magnetic field due to the current is equal to the magnetic field of earth.
[tex]B_{E}=B_{I}=\dfrac{\mu_{0}NI}{2r}[/tex]
We need to calculate the current in the loop
Using formula of magnetic field
[tex]B=\dfrac{\mu_{0}NI}{2r}[/tex]
[tex]I=\dfrac{2rB}{\mu_{0}N}[/tex]
Put the value into the formula
[tex]I=\dfrac{2\times9.4\times10^{-2}\times0.7\times10^{-4}}{4\pi\times10^{-7}\times1}[/tex]
[tex]I=10.5\ A[/tex]
Hence, The current in the loop is 10.5 A.
HELP PLS!!
In a full/angled projectile, Vty is equal to the inverse of viy
O True
O False
Which pair of objects would be most strongly attracted to each other?
A. A positively charged particle and a negatively charged particle
B. Two positively charged particles
C. Two negatively charged particles
D. A negatively charged particle and a neutral particle
Answer:
Its A
Explanation:
Just did the quiz
The pair of objects that should be strongly attracted is option A. A positively charged particle and a negatively charged particle.
Pair of objects that are most strongly attracted:When there is a positively charged particle & the negatively charged particle so due to this it should be strongly attracted. Also, when there are two positively charged particles, or a negative one or a include negative one or neutral one so it should not be strongly attracted. Therefore, the option A is correct.
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Compare and contrast the CONFLICT (choose one) in the short story you read with the elements appearing in The Watsons Go to Birmingham—1963. Explain how they are similar or different in a few sentences.
Answer:
they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta
Explanation:
use in your own words teachers know when your not trust me.
A dog has a mass of 60kg and an acceleration of 2m/s/s. What is the force of the dog?
A spring gun is able to launch a 7.0 gram marble to a vertical height of 22 mmeasured from the compressed point of the marble). If the spring iscompressed 8.0 cm from its relaxed state, what will be the spring constant
Answer: Spring constant = 472N/m
Explanation:
The change in gravitational potential energy by the spring is given as = mgh
where m= 7.0 g = 7 X 10 -3kg
g= 9.8m/s
h= 22m
Gravitational potential energy= mgh
= 7.0 x 10^-3 X 9.8 x 22 = 1.5092 J
Remember that change in gravitational potential energy by the spring =elastic potential energy
Therefore, Potential energy P. E = 1/2 K x²
where K= CONSTANT
x= 8.0
2 X 1.5092 J / (8.0 X 10^-2)²= 471.625 ROUNDED TO 472 N/m
A water balloon launcher with a mass of 2.2 kg is suspended on a wire. It fires a 0.85 kg balloon to the north at a velocity of 13.0 m/s. What is the resulting velocity of the launcher if the net force on the launcher is equal to the reaction force?
Answer:
5.0 m/s south
Hope this Helps!
Answer:
5.0 m/s south
Explanation: