Answer:
acceleration equal to 0.5 m / s
Explanation:
since 20 x 0.5 = 10
A shell is fired from the ground with an initial speed of 1.60 × 10^3 m/s (approximately five times the speed of sound) at an initial angle of 68.0◦ to the horizontal. The acceleration of gravity is 9.81 m/s^2 Neglecting air resistance, find the shell's horizontal range? Answer in units of m
Answer:
181276.802 m
Explanation:
The computation of the horizontal range of the shell is shown below:
As we know that
range = u × usin2θ ÷ g
where,
u = 1600 m/s
θ = 68
g=9.81
NowPut these values to the above formula
So, the value of the horizontal range of the shell is
R = 181276.802 m
= 181276.802 m
We simply applied the above formula and the same is to be considered
Two observers times the motion of a car from one place to another. The first observer's clock read 262 seconds at the start and 375 seconds at the end of the car's motion. The second observer's clock read -86 seconds at the start. What did it read at the end?
Answer:
The time of the second observer at the end of the car motion is 27 s
Explanation:
Initial time of the first observer, t₁ = 262 s
final time of the first observer, t₂ = 375 s
The time of the car motion, t = t₂ - t₁
t = 375 s - 262 s
t = 113 s
Initial time of the second observer, t₁ = -86 s
final time of the second observer, t₂ = ?
The time of the car motion, t = t₂ - t₁
113 = t₂ - t₁
113 = t₂ - (-86)
113 = t₂ + 86
t₂ = 113 - 86
t₂ = 27 s
Therefore, the time of the second observer at the end of the car motion is 27 s
A car of mass 1500kg is brought to rest from a velocity of 25m/s by a constant force of 3,000N. Determine the change in momentum produced by the force and the time it takes for the car to come to rest.
change in momentum = 37,500kg.m/s in the opposite direction
time it takes = 12.5s
u=25m/s
v=0m/s
F=ma
-3000=(1500)(a)
-3000÷1500=a
a= -2m/s^2
Change in momentum=mv-mu
(1500)(0)-(1500)(25)
= -37,500kg.m/s
relax the( - ) represents direction
v=u-at
0=25-2t
2t=25
t= 12.5s
What is the mass(M) of a 330 N television? gon earth is = 9.8 m/s2
Explanation:
Well, it depends...
Explanation:
Weight is given by the formula,
W
=
m
g
W
is the weight of the object in newtons
m
is the mass of the object in kilograms
g
is the gravitational acceleration
On Earth,
g
=
9.8
m/s
2
. So, the object's weight on Earth will be:
W
=
50
kg
⋅
9.8
m/s
2
=
490
N
Can you tell what the object's weight will be on another planet, such as Mars?
Answer: 34 kilograms (approximate)
==================================
Work Shown:
F = m*g
330 newtons = (m kg)*(9.8 m/s^2)
330 = m*9.8
330/9.8 = m
m = 33.673469387755
m = 34 kilograms
I'm rounding to 2 significant figures because 330 has 2 sig figs and 9.8 has 2 sig figs. When dividing we pick the smaller of the sig fig count to determine how many sig figs to round to.
How a killer whale causes deforestation of the seabed
We can confirm that the primary reason that killer whales cause deforestation of the seabed is due to a domino effect caused by one of their food sources.
How do whales cause deforestation?Killer whales are carnivorous organisms. Among their prey are sea otters. This new source of food for killer whales began to cause a domino effect in that the decline in sea otter population caused the primary meal of the otters, that being sea urchins, to increase greatly in population. Given that the sea urchins feed on kelp, this contributes to the deforestation of the sea bed and the lack of kelp.
Therefore, we can confirm that killer whales cause deforestation of the seabed in an indirect manner by contributing to the increase in sea urchin populations that leads to deforestation of the seabed.
To learn more about ecosystems visit:
https://brainly.com/question/1673533?referrer=searchResults
"Angel Falls in southeastern Venezuela is the highest uninterrupted waterfall in the world, dropping 979 m (3212 ft). Ignoring air resistance, it would take 14 s for the water to fall from the lip of the falls to the river below. If the water lands 50 m from the base of the verti-cal cliff, what was its horizontal speed at the top"
Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is [tex]v = 3.57 \ m/s [/tex]
Explanation:
From the question we are told that
The depth is [tex]d = 979 \ m[/tex]
The time taken for water to fall to below the river is [tex]t = 14 \ s[/tex]
The horizontal distance from the base of the vertical cliff is [tex]h = 50 \ m[/tex]
Generally the horizontal speed at the top is mathematically represented as
[tex]v = \frac{h}{t}[/tex]
=> [tex]v = \frac{ 50}{ 14 }[/tex]
=> [tex]v = 3.57 \ m/s [/tex]
A statue and a coin are made out of exactly the same materials. Which property would you claim will likely be the same for both the statue and the coin?
mass
reactivity
volume
shape
I believe that the correct answer is reactivity.
What is the electrical force between q2 and q3? Recall that k = 8. 99 × 109 N•meters squared over Coulombs squared. 1. 0 × 1011 N –1. 1 × 1011 N –1. 6 × 1011 N 1. 8 × 1011 N.
Force on the particle is defined as the application of the force field of one particle on another particle. the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.
What is electric force?Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.
The electric force in the second case will be the same as in the first case. Therefore the force on the particle will be the same.
[tex]\rm F= K\frac{q_2q_3}{r^2}[/tex]
[tex]\rm F= 9\times 10^9 \times \frac{1.6 \times 10^{-13}\times 1.6\times10^{-13}}{(0.5)^2}[/tex]
[tex]\rm F= - 1. 1 \times 10^{11 }N[/tex]
Hence the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.
To learn more about the electric force refer to the link;
https://brainly.com/question/1076352
3) (a) A piece of brass weighs 175g in air and 150 g when fully immersed in water. The
density of water is 1.0gcm-3.
What is the volume of the brass piece?
(ii) (ii) Why does the brass piece weigh less in water?
Answer:
[tex]\rainbow{\rule{10000pt}{900000pt}}[/tex]
10 20 30 40 50 60 70 80 90 100 110 120 130. Which end of the scale (left or right) has the loudest sounds?
Answer:
I'm From Philippines
I hope we can be friends.
A room measures 3 meter wide, 4 meters long, and 2.5 meters high. a. What is the surface area of the room in square meters (including the ceiling and the floor)? b. If a certain paint covers 25 square meters per gallon, how many gallons will it take to paint the walls and ceiling, but not the floor?
Answer:
Explanation:
The total surface area of the room is expressed as;
TSA = 2(LW+LH+WH)
L is the length
W is the width.
H is the height
Given
L = 4m
W = 3m
H = 2.5m
.TSA = 2{4(3)+3(2.5)+4(2.5)}
TSA = 2(12+7.5+10)
TSA = 2(29.5
TSA = 59m²
TSA (without the floor) = 2(3(2.5)+4(2.5))
TSA = 2(7.5+10)
TSA = 2(17.5)
TSA = 35m²
If a certain part covers 25 square meters per gallon, the amount of gallon needed will be 35m²/25m²
= 1.4gallons
If wooden ball has a mass of 2.4g and a volume of 6cm3. What is it’s density
Answer:
0.4 g/cm^3. is the answer of this question..Thank you ☺️☺️
Answer:
0.4 g/cm^3
Explanation:
As density=m/v
=2.4/6
=0.4g/cm^3
25 POINTS will mark brainliest if correct
The radius of Neptune is 2.48 x 107m. The radial acceleration of an object at the equatorial surface of
The radius of Neptune is 2.48*10^7m. The radical acceleration of an object at the equatorial surface of Neptune is 0.292 m/s2. What is Neptune's period of rotation in seconds?
A.18,000 s
B.1200 s
C.58,000 s
D.9200 s
Answer: The answer is C. 58,000
Explanation:
a meteor is falling towards earth with 4m/s^2 acceleration due to gravity. if mass and radius of the earth are 6*10^24kg and 6.4*10^6m respectively, find the height of meteor.
Answer:
3.6×10⁶ m
Explanation:
g = GM / R²
4 m/s² = (6.67×10⁻¹¹ m³/kg/s²) (6×10²⁴ kg) / (6.4×10⁶ m + h)²
(6.4×10⁶ m + h)² = 1.00×10¹⁴ m²
6.4×10⁶ m + h = 1.00×10⁷ m
h = 3.6×10⁶ m
Which of the following is usually tax exempt?
Answer:
Explanation:
whatever expenses a business might have like
repairs
buying equipment
employees
How long would it take for a car to travel 3200 km at 80 km/hr East?
Answer:
3.75 hr.Explanation:
distance = 300 km
velocity = 80 km/hr.
find: time
time = distance
velocity
= 300 km
80 km/hr.
= 3.75 hr.
On Earth, a board is supported near the ends by Laney and Mia. If Laney exerts 52N upward and
Mia exerts 35 N upward, find the mass of the board if it is in Translational Equilibrium. (Type
answer with 2 significant digits)
Answer:
??? this makes no sense give more info
Explanation:
Select the correct answer. What can be found by counting the number of troughs per second in a wave diagram? A. Amplitude B. Direction C. Frequency D. Speed E. Wavelength.
Option C is correct. Frequency can be found by counting the number of troughs per second in a wave diagram.
What is the frequency?Frequency is defined as the number of repetitions of waves occurring in 1 second. It can also be said that Frequency is defined as the number of cycles completed in 1 second.
Frequency is given by the formula as,
[tex]\rm f=\frac{1}{t}[/tex] Hz.
Hence option C is correct. Frequency can be found by counting the number of troughs per second in a wave diagram.
To learn more about the frequency refer to the link;
https://brainly.com/question/4393505
Answer:
C is correct
Explanation:
heeelppp what quantity measures the amount of matter in an object
A . Volume
B . Temperature
C . Mass
D . Density
Answer:
Your answer is MASS.
Explanation:
Volume measures how much space an item takes up.
Temperature measures how hot or cold something is.
Density is the degree of compactness of a substance.
Mass measures the amount of matter in an object.
a 5.00 kg mass is at the origin. a 7.00 kg mass is located at x = 0.600 m, and a 4.00 kg mass is at x = 0.900 m. What is the net gravitational force on the 5.00 kg mass?
Answer:
Given,
A 5 kg mass is at the origin. A 7 kg mass is located at x = 0.6 m and a 4 kg mass is at x = 0.9 m.
To find,
the net gravitational force on the 5 kg mass.
we know, from Newton's gravitational law, two massive bodies attract each other by a force , F = Gm₁m₂/r²
so, 7 kg mass attracts 5kg mass by the force, F₁ = G(7kg)(5kg)/(0.6)²
= 35G/0.36 = 97.22G
similarly, 4kg mass attracts 5kg mass by the force, F₂ = G(4kg)(5kg)/(0.9)²
= 20G/0.81 = 24.69G
net gravitational force = F₁ + F₂
= 97.22G + 24.69G
= 121.91G
we know, G = 6.67 × 10¯¹¹ Nm²/Kg²
F = 121.91 × 6.67 × 10¯¹¹
= 813.1397 × 10¯¹¹ N
= 8.131397 × 10^-9 N ≈ 8.13 × 10^-9 N
Therefore net gravitational force acting on mass 5kg is 8.13 × 10^-9 N
The tidal bulge on earth follows the movement of the:.
Answer:
The moon and sun cause tidal bulges to move around the planet in perfect sync and unison.
Explanation:
However, do not consider these to be large oceans of water moving in relation to continents. Only the fluctuations in water level—the water's surface profile—follow the moon's and sun's positions in the sky.
What is the relationship between current and power in a circuit?
O A. A circuit that has more current will have more power.
B. The amount of current in a circuit has no relation to its power.
C. A circuit that has more current will have less power.
D. The amount of current a circuit has is equal to its power.
Answer: A
Explanation: a p e x
Answer:
the guy above is right, it is A
Explanation:
A p E x
Which form of electromagnetic light can be used for cooking food?
A) X-rays
B) Ultra-violet (UV) light
C) Gamma Rays
D) Microwaves
Answer:
Microwaves (D
Explanation:
I'm not sure! But I think it's Microwaves!
The pressure resulting from a force of 50 N exerted over an area of 5 m² is
P = F/A
P = 50/5
P = 10 Pa
Hope it helps :)
PLEASE HELP!!!!!!!!!!
Because the boy is slowing down, his velocity decreases over time. Use the final and starting velocities in the data table to calculate his average acceleration. Use the equation , where t is time, ux is initial velocity, vx is final velocity, and ax is acceleration. Use time t = 1.6 seconds, which is calculated using values from the table (1.8 − 0.2).
Answer:
The answer should be 0.825
Explanation:
I took like 20 mins to figure this out lol
please help me guys i really dont understand this one
Answer:
Density is the degree of compactness in an object.
Explanation:
S.I unit is kg/m3
Answer:
The mass of the paraffin is 70g
Explanation:
You have to subtract the mass of the empty beaker from that of the empty beaker and paraffin oil to get the mass of the paraffin oil
A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. On the rough section, the coefficient of friction is not constant, but starts at 0.100 at P and increases linearly with distance past P, reaching a value of 0.600 at 12.5 m past point P.
(a) Use the work energy theorem to find how far this box slides before stopping.
(b) What is the coefficient of friction at the stopping point?
(c) How far would the box have slid iff the friciton coefficient didn't increase, but instead had the constant value of .1?
Answer:
(a). The value of distance is 5.043 m
(b). The coefficient of friction is 0.302.
(c). The distance is 10.33 m
Explanation:
Given that,
Speed = 4.50 m/s
Minimum friction = 0.1
Maximum friction = 0.6
Distance = 12.5 m
We know that,
The newton's second law
[tex]F_{net}=-\mu N[/tex]
[tex]F_{net}=-\mu mg[/tex]
We know that,
The frictional coefficient is directly proportional to the distance.
We need to calculate the value of coefficient
Using general equation of frictional coefficient
[tex]\mu(x)=Ax+B[/tex]....(I)
At x= 0,
[tex]\mu(0)=0+B[/tex]
Put the value into the formula
[tex]B=0.1[/tex]
Now for value of A
Put the value in equation (I)
[tex] 0.6=A12.5+0.1[/tex]
[tex]A=\dfrac{1}{25}[/tex]
Put the value in equation (I)
[tex]\mu(x)=\dfrac{x}{25}+0.1[/tex]......(II)
We need to calculate the work done
Using work energy theorem
[tex]W=\int_{0}^{x}{F_{net} dx}[/tex]
[tex]W=\int_{0}^{x}{-\mu_{x}mgdx}[/tex]
[tex]W=-mg\int_{0}^{x}{(\dfrac{x}{25}+0.1)dx}[/tex]
[tex]W=-mg(\dfrac{x^2}{50}+0.1x)[/tex]....(III)
(a). We need to calculate the distance
Using difference of kinetic energy
[tex]W=\dfrac{1}{2}m(v_{f}^2-v_{0}^2)[/tex]
[tex]-mg(\dfrac{x^2}{50}+0.1x)=\dfrac{1}{2}m(0-(4.50)^2)[/tex]
[tex]\dfrac{x^2}{50}+0.1x=1.013[/tex]
[tex]x^2+5x=50.65[/tex]
[tex]x^2+5x-50.65=0[/tex]
[tex]x=5.043\ m[/tex]
(b). We need to calculate the coefficient of friction at the stopping point
Using equation (III)
[tex]\mu(5.043)=\dfrac{5.043}{25}+0.1[/tex]
[tex]\mu(5.043)=0.302[/tex]
(c). We need to calculate the distance
Using formula work done
[tex]W=-fx_{2}[/tex]
[tex]W=-\mu mgx_{2}[/tex]
[tex]-\dfrac{1}{2}mv_{0}^2=-\mu mgx_{2}[/tex]
[tex]x_{2}=\dfrac{v_{0}^2}{2\mu g}[/tex]
Put the value into the formula
[tex]x_{2}=\dfrac{(4.5)^2}{2\times0.1\times9.8}[/tex]
[tex]x_{2}=10.33\ m[/tex]
Hence, (a). The value of distance is 5.043 m
(b). The coefficient of friction is 0.302.
(c). The distance is 10.33 m
Determin the acceleration that results when an object starts at rest moves a distance and it’s final speed is 49 m/sec
(show work please and equation used)
(also don’t understand this question at all)
Answer:
The answer is below
Explanation:
The question is not complete since the distance is not given. But let us assume the object moves a distance of 100 m.
From newton's law for a moving body, we get the equation:
v² = u² + 2as
Where v is the final velocity of the object, u is the initial velocity of the object, s is the distance moved by the object and a is the acceleration of the object.
Given that the object is initially at rest, hence u = 0, final speed (v) = 49 m/s, distance (s) = ?
v² = u² + 2as
Substituting:
49² = 0² + 2as
2401 = 2as
a = 2401 / 2s
Substituting s = 100 m
a = 2401 / (2*100)
a = 12 m/s²
Sarah throws a ball directly upward at the edge of a cliff with a starting velocity of 3.0 \, \dfrac{\text m}{\text s}3.0 s m 3, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. It lands on the ground with a final speed of 6.0 \, \dfrac{\text m}{\text s}6.0 s m 6, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. How long is the ball in the air?
Answer:
t = 0.3 seconds
Explanation:
Given that,
Initial velocity of a ball, u = 3 m/s
Final velocity of the ball, v = 6 m/s
We need to find the time for which the ball is in the air. As the ball was directed upwards, it would mean that the acceleration of the ball is -g or -9.8 m/s².
Using the first equation of motion to find the time. So,
v=u+at
or
v=u-gt
[tex]t=\dfrac{v-u}{-g}\\\\t=\dfrac{6-3}{-9.8}\\\\t=-0.3\ s[/tex]
or
t = 0.3 seconds
So, the ball is in the air for 0.3 seconds.
Answer: t= 0.92 s
Explanation: the formula you use it’s v=vinitial + at and you rearrange it to make (t= vfinal - vinitial /a) using this you get 0.92s
7. A sprinter accelerates from rest to a velocity of 12 m/s in the first 6
seconds of the 100-meter dash.
What is the acceleration in the first 6 seconds?
b. How far does the sprinter travel during the first 6 seconds?
Answer: a) 36 m/s
B) 64 m
Explanation:
A) d= ½ (Vi+Vf)t
d= ½ (0+12)6
d= 36 m/s
B) 100 – 36 = 64