Answer:
F = 2700 N
Explanation:
Given that,
Mass of a car, m = 3000 kg
Initial velocity, u = 15 m/s
Final velocity, v = 60 m/s
Time, t = 5 s
We need to find the net force on the car during this time . The net force on an object is given by the product of mass and acceleration. So,
F = ma
a is acceleration, a = (v-u)/t
[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{3000\times (60-15)}{5}\\\\F=27000\ N[/tex]
So, 27000 N is the net force on the car during this time .
8. When are children MOST likely to begin using words to express meaning?
what is an electric current ?
Answer:
An electric current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor or space. It is measured as the net rate of flow of electric charge through a surface or into a control volume.
Explanation:
An object is placed at 0 on a number line. It moves 3 units to the right, then 4 units to the left and then 6 units to the
right. What is the displacement of the object?
1
5
7
13
Answer:
5
Explanation:
when you move 3 unit to the right your object is at 3.then when you move 4 units to the left your object is at -1 because
a number line also has negative numbers.Then if you move 6 units your object is at 5
If a system does 63.0 kJ of work on its surroundings and releases 105 kJ of heat, what is the change in the internal energy of the system, in kJ
Answer:
-42 kJ
Explanation:
Q = ΔE + W
W = -63.0 kJ
Q = -105 J
so -105 = ΔE + (-63)
ΔE = 63 - 105 = -42 kJ
A system does the work of 63 kJ and releases 105 kJ of heat, then the change in internal energy will be -42 kJ.
What is heat?When a thermodynamic system's boundary is crossed by a form of energy due to a temperature differential, that form of energy is referred to as heat. Heat is not present in thermodynamic systems.
In the right-hand image, a metal bar is "conducting heat" from its hot end to its cold end. However, if the metal bar is thought of as a thermodynamic system, then the energy moving within the metal bar is what is being described as "conducting heat" instead is called internal energy.
Q = Δ E + W
W = -63.0 kJ
Q = -105 J
Then,
-105 = Δ E + (-63)
ΔE = 63 - 105 = -42 kJ
Therefore, the change in internal energy will be -42 kJ.
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students in a lab are doing experiments involving the motion of objects. they pull an object across a table and use a detector to measure the acceleration of the object for the entire time they pull it. The graph shows the data from the detector, with four segments labeled
Based on the acceleration-time graph, the segments of the motion in which the object experience a net force are both 1 and 3.
What is Newton's Second Law of Motion?Newton's Second Law of Motion states that the acceleration of an object is inversely proportional to its mass and directly proportional to the net force acting on the physical object.
How to calculate the net force.Mathematically, the net force acting on the car is given by Newton's Second Law of Motion:
[tex]F=ma =\frac{m(v\;-\;u)}{t}[/tex]
Where:
m is the mass.t is the time.u is the initial velocity.u is the final velocity.a is the acceleration.Note: Mass represents the conversion factor between net force and acceleration on an acceleration-time graph.
Based on the acceleration-time graph, the segments of the motion in which there is a net force acting on the object are both 1 and 3 while it experienced constant acceleration at segment 2 and 4.
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If a satellite is launched to a distance of 1000000m above the earth’s surface, what is the gravitational field strength on the satellite from the earth?
Explanation:
A 620 kg satellite above the earth's surface experiences a gravitational field strength of 4.5 N/kg.
Knowing the gravitational field strength at Earth’s sur-
face and Earth’s radius, how far above Earth’s surface
is the satellite? (Use ratio and proportion.)
HELP PLEASE!!!!!!!!!
Answer:
2.00×[tex]10^{8}[/tex]
Explanation:
3×[tex]10^{8}[/tex]/1.5
=2.00×[tex]10^{8}[/tex]
A 30 kg child is in a school play. She is standing on the ground at rest but she is attached to a rope with a 5 kg mass attached. A 45 kg child takes off the 5 kg mass and jumps from a height in order to hoist the 30 kg child off the ground. What is the acceleration that both children have together?
:)
PLEASE HELP AND I WILL GIVE U BRAINLY
a south african white rhinoceros weighs 22658 N and has a top speed of 14 m/s. a bullet from a barrett 0.50 caliber rifle has a weight of 0.4 N and travels with a velocity of 884 m/s. Which object has more momentum in its motion?
Answer:
it is the white rhinoceros because it weighs more then the bullet all though the bullet travels faster once the rhino gets running it is hard to stop because it weighs so much
Is the moon always showing up at the same time?
Answer:
the moon showing up at same time as what
Explanation:
???
an Alaskan rescue plane traveling 59 m/s drops a package of emergency rations from w height of 168 m to a stranded party of explorers.
the accelerating of gravity is 9.8 m/s^2.
where does the package strike the ground relative to the point directly below where it was released? answer units of m
A 3520 kg truck moving north at
18.5 m/s makes an INELASTIC
collision with an 1480 kg car
moving east. After colliding, they
have a velocity of 13.6 m/s at
72.6°. What was the initial velocity
of the car?
Answer: 13.7 m/s
Explanation:
[tex]$$Mass of $T r u c k$$ \ \left(m_{T}\right)=3520\mathrm{kg}[/tex]
[tex]V_{T_{y}}=18.5 \mathrm{~m} / \mathrm{s} \\&V_{\text {Tix }}=0 \mathrm{~m} / \mathrm{s}[/tex]
[tex]\mathrm{Mass \ of \ car \ (m_c) = 1480 \ kg}[/tex]
[tex]V_{c i x}=\text { ? ; } V_{c i y}=0 \mathrm{~m} / \mathrm{s}[/tex]
[tex]Final \ velocity $\left(V_{f}\right)=13.6 \mathrm{~m} / \mathrm{s} \ \ \theta=72.6$[/tex]
[tex]v_{f x}=v_{f} \cos \theta=(13.6) \cos 72.6=4.067 \mathrm{~m} / \mathrm{s}[/tex]
[tex]\mathrm {Using \ the \ conservation \ of \ momentum \ along\ the \ $x$-axis}[/tex]
[tex]\begin{aligned}m_{T} v_{T i x}+m_{c} v_{c_{i x}} &=\left(m_{T}+m_{c}\right) v_{f x} \\0+(1480) V_{c_{i x}} &=(3520+1480)(4.067) \\(1480) v_{c i x} &=20335 \\V_{c i x} &=13.7 \mathrm{~m} / \mathrm{sec}\end{aligned}[/tex]
Answer:
13.7 m/s
Explanation:
list clown the advantages and disadvantages
of the micrometer screw guage
Caliper?
over Vernier
Answer:
Micrometers provide very accurate measurements
The micrometer is one of the most accurate types of measuring device.
Most micrometers can measure up to 0.001mm or 0.0001 inches.
-------
Ratchet speeder helps to provide reliable measurements
The ratchet speeder/stop mechanism ensures that uniform pressure is applied to the thimble so that measurements are reliable and repeatable.
-------
Explanation:
what is digital vernier calliper
Answer:
a precision instrument for taking extremely precise measurements. It's the newest advancement in caliper technology.
Explanation:
The average standard rectangular building brick has a mass of 3.10 kg and dimensions of 225 m x 112 m x 75 m. The gravitational field constant g = 9.8 N/kg. Calculate the maximum pressure created by brick
Answer:
[tex]P=3.61\times 10^{-3}\ Pa[/tex]
Explanation:
Given that,
Mass of a brick, m = 3.1 kg
The dimensions of the brick 225 m x 112 m x 75 m
We need to find the maximum pressure created by the brick. We know that, the force acting per unit area is called pressure exerted. It is given by the formula as follows :
[tex]P=\dfrac{F}{A}[/tex]
F = mg
A = area with minimum dimensions i.e. 112 m x 75 m
Pressure is maximum when the area is least.
So,
[tex]P=\dfrac{mg}{A}\\\\P=\dfrac{3.1\times 9.8}{112\times 75}\\\\P=3.61\times 10^{-3}\ Pa[/tex]
So, the maximum pressure created by brick is [tex]3.61\times 10^{-3}\ Pa[/tex].
1. A surfboarder rides a wave for 23.7 m at a constant rate of 4.1 m/s. How long did his trip
take?
Answer:
His trip took 5.78 seconds
Explanation:
23.7m divided by 4.1m/s = 5.78048780488
Match each description with the correct graph
Answer: 1. B, 2. A, 3. C, 4. D
Answer:
1 = B
2 = A
3 = C
4 = D
I hope this helps
Please help 50 points: A ball is thrown straight up, from 3 m above the ground, with a velocity of 14m/s. When does it hit the ground? Use vertical motion formula.
Projectile Motion for an Object Launched at an Angle
When an object is projected from rest at an upward angle, its initial velocity can be resolved into two components. These two components operate independently of each other. The upward velocity undergoes constant downward acceleration which will result in it rising to a highest point and then falling backward to the ground. The horizontal motion is constant velocity motion and undergoes no changes due to gravity.The analysis of the motion involves dealing with the two motions independently.
Explanation:
and Thais
Projectile Motion for an Object Launched at an Angle
When an object is projected from rest at an upward angle, its initial velocity can be resolved into two components. These two components operate independently of each other. The upward velocity undergoes constant downward acceleration which will result in it rising to a highest point and then falling backward to the ground. The horizontal motion is constant velocity motion and undergoes no changes due to gravity.The analysis of the motion involves dealing with the two motions independently.
Projectile Motion for an Object Launched at an Angle
When an object is projected from rest at an upward angle, its initial velocity can be resolved into two components. These two components operate independently of each other. The upward velocity undergoes constant downward acceleration which will result in it rising to a highest point and then falling backward to the ground. The horizontal motion is constant velocity motion and undergoes no changes due to gravity.The analysis of the motion involves dealing with the two motions independently.
An object is swung in a horizontal circle on a length of string
that is 0.93 m long. Its acceleration is 26.36 m/s? What is the
time it takes the object to complete one horizontal circle?
Answer: either 3.20s or 1.18s
Explanation:
the diagram below shows the movement of matter in a portion of the water cycle.
what does p most likely represent
rivers
lakes
infiltration
precipitation
Answer:
Infilteration
Which one these statements describes how to typically tell the difference
between a synthesis (S) reaction and a decomposition (D) reaction?
Tap to select or deselect answer.
An S reaction has one product and two or more reactants and a D reaction
has one reactant and two or more products.
An Sreaction has one reactant and two or more products and a D reaction
has one product and two or more Reactants.
An S reaction has oxygen as a reactant and a Dreaction has oxygen as a
product
An S reaction has oxygen as a product and a D reaction has oxygen as a
reactant
Answer: An S reaction has one product and two or more reactants and a D reaction has one reactant and two or more products.
Explanation:
In a synthesis reaction, two or more simple molecules or substances interact to form a more sophisticated product. (General equation: A + B → AB)
When one reactant decomposes into two or more products, this is known to as a decomposition reaction. (General equation: AB → A + B)
what is kinetic energy
Answer:
energy which a body possesses by virtue of being in motion.
Explanation:
hope this help
pls mark as brainliest
A disc of radius wich uniform distribution of mass rotates about on axis perpendicular to it's plane at the rim with angular speed, the moment of inertia of the disc about the axis through the centre is 1/2MR*R, what's the kinetic energy of the disc?
Answer:
Explanation:
M3U1 Pre-Assessment
Reflection Journal - Thursday
A delivery drone uniformly slowed down from 45 m/s to 23 m/s in a straight line over a 30-minute duration within a 16-
kilometer radius of its warehouse. What was the magnitude of its average acceleration?
Answer: The Answer is A 1.2 x 10−2 m/s2
The average acceleration of the delivery drone as it slowed down was -0.012 m/s².
What is average acceleration?The average acceleration is defined as the average rate of change of velocity with respect to time. Mathematically -
a = Δv/Δt
Given is a delivery drone that uniformly slows down from 45 m/s to 23 m/s in a straight line over a 30-minute duration.
We can write -
[u] = 45 m/s
[v] = 23 m/s
Δt = 30 min = 30 x 60 = 1800 sec
We can write the average acceleration as -
a = Δv/Δt
a = (23 - 45)/1800
a = - 0.012 m/s²
Therefore, the average acceleration of the delivery drone as it slowed down was -0.012 m/s².
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You walk out to the edge of a 2,384 m. cliff and you toss a rock down to watch it drop. What is the rocks acceleration?
Answer:
9.8m/s
Explanation:
Wherever you're on the earth's surface, gravitational force remains the same and it does not matter either its a rock that's being tossed or a sheet of paper, the acceleration remains 9.8m/s only. *Condition applied=air resistance neglected
A framed picture hangs from two cords attached to the ceiling. Which shows the correct free body diagram of the hanging picture?
Answer: the answer is in the picture
Explanation:
Answer:
C
Explanation:
i did the test and got 100 :)
Help please it’s urgent
Answer:
Explanation:
The circuit is incomplete since it is a series circuit
If a person weighs 692 N on earth and 5320 N on the surface of a nearby planet, what is the acceleration due to gravity on that planet?
Answer:
Approximately [tex]75.3\; \rm m \cdot s^{-2}[/tex]. (Approximately [tex]7.69\, g[/tex], where [tex]g[/tex] denotes the acceleration due to gravity on the surface of the earth.)
Explanation:
The weight of an object is the size of the overall gravitational pull on that object. If the mass of this person is [tex]m[/tex], the weight of this person at a point with gravitational acceleration [tex]g[/tex] would be [tex]m \cdot g[/tex].
Let [tex]g_{0}[/tex] denote the gravitational acceleration on the surface of the earth. The weight of this person on the surface of the earth would be [tex]m \cdot g_0[/tex].
Let [tex]g_1[/tex] denote the gravitational acceleration on the surface of that "nearby planet". The weight of this person on the surface of that planet would be [tex]m \cdot g_1[/tex].
According to the question:
Weight of this person on the surface of the earth: [tex]692\; \rm N[/tex]. Therefore, [tex]m \cdot g_0 = 692\; \rm N[/tex].Weight of this person on the surface of that nearby planet: [tex]5320\; \rm N[/tex]. Therefore: [tex]m \cdot g_1 = 5320\; \rm N[/tex].Take the quotient of these two equations:
[tex]\displaystyle \frac{m\cdot g_1}{m \cdot g_0} = \frac{5320 \; \rm N}{692\; \rm N}[/tex].
[tex]\displaystyle \frac{g_1}{g_0} \approx 7.6879[/tex].
[tex]g_1 \approx 7.6879 \, g_0[/tex].
In other words, the gravitational acceleration on the surface of that nearby planet is approximately [tex]7.6879[/tex] times the gravitational acceleration on the surface of the earth.
The gravitational acceleration on the surface of the earth is approximately [tex]g \approx 9.8\; \rm m \cdot s^{-2}[/tex]. Therefore, the gravitational acceleration on the surface of that planet would be approximately [tex]7.6879\, g \approx 7.8679 \times 9.8\; \rm m \cdot s^{-2} \approx 75.3\; \rm m \cdot s^{-2}[/tex].
A pulley in the form of a uniform disk)
with mass 63 kg and a radius 20 cm is at-
tached to the ceiling in a uniform gravita-
tional field and rotates with no friction about
its pivot. The masses are connected by a
massless inextensible cord and T1, T2, and T3
are magnitudes of the tensions.
T3
20 cm
T2
63 kg
1
39 kg
T1
3.8 m
15 kg
2
What is the magnitude of the tension T?
The acceleration due to gravity is 9.8 m/s?.
Assume up is positive.
Answer in units of N.
For a pulley in the form of a uniform disk with a mass of 63 kg and a radius of 20 cm, the magnitude of the tension T is mathematically given as
T1=292.4N
What is the magnitude of the tension T?For 20kg
T1=196+20a
For
33kg
T2=323.4-33a
Generally, the simultaneous equation is mathematically given as
T2-T1=(323.4-33a)-(196+20a)
Hence
127.4-53.a
a=1.67m/s^2
T1=196+20(1.67)
T1=292.4N
In conclusion, the magnitude of the tension T
T1=292.4N
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Suppose a radio signal travels from Earth and through space at a speed of 3.0 × 108 m/s. How far into space did the signal travel during the first 33.3 minutes?
For a radio signal that travels from Earth and through space at a speed of 3.0 × 108 m/s, The distance into space the signal traveled during the first 33.3 minutes is mathematically given as
d=647352m
How far into space did the signal travel during the first 33.3 minutes?Generally, the equation for the distance is mathematically given as
d=v*t
Where
t=33.3min
t=33.3*60
t=1998sec
Therefore
d=3.0 × 108 m/s *1998sec
d=647352m
In conclusion, distance is
d=647352m
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