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Varvara68 [4.7K]
4 years ago
15

You build a 2.0-kg mound of mashed potatoes on a kitchen scale (which is a spring scale). You then use a large spoon to press on

the top of the mound at a constant force of 3.0 N. While you are doing this, the scale reading increases by 5.0%. (a) What is the acceleration of the center of mass of the potatoes? (b) Describe what happens to the shape of the mound and the scale reading as time goes by.

Physics
2 answers:
igor_vitrenko [27]4 years ago
8 0

Answer:

a) a=1m/s^2 downwards

b) The mound will be reduced in height (the acceleration is downwards) and will begin to grow sideways. The reading will remain the same until the spoon reaches the scale surface.

Explanation:

On the mound, the forces acting before the spoon are:

N1 - m*g = 0   Where N1 is equivalent to the reading of the scale.

N1 = 2*10 = 20N

After the spoon starts pressing, this value increases by 5%:

N2 = 20*105% = 21N  So, at this point the forces on the moud are:

N2 - m*g - F = m*a   Solving for a:

a = -1m/s^2   This result tells us that the center of mass of the potatoes is moving downwards. Since it cannot move because of the scale, it has no other option than to grow sideways as it is pushed.

marshall27 [118]4 years ago
5 0

Answer:

a= -1.01m/s^2

Explanation:

A) We define all the variables that are given,

In this way

Acceleration = a

m = 2.0Kg

F = 3.0N

We obtain the Normal assuming 5% of the potato reading scale, so

N = (2 + 2 * 5\%) * 9.8

N = 20.58N

We add forces to reach equilibrium, and you get

N-F-mg = ma

** The notation of the symbols is done according to the notation of the graph

So,

a = \frac {N-F-mg} {m}

a = \frac {20.58-3-2 * 9.8} {2}

a = -1.01m / s ^ 2

B) When making pressure it is understood that the shape takes the perpendicular direction of the direction of the force that is performed. If the pressure remains constant, the scale will also increase in its measurement.

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3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

4 0
3 years ago
Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0
3 years ago
How much force is required to accelerate a 800kg car by 15 m/s2
Tom [10]

Answer:

force=12000

Explanation:

F=m*a aka force equals mass times acceleration so 800*15=12000

3 0
2 years ago
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Some miners wish to remove water from a mine shaft. A pipe is lowered to the water 90 m below, and a negative pressure is applie
WITCHER [35]

Answer:

Explanation:

Given

Pipe is lowered to the water h=90\ m

Negative Pressure is applied to raise the water

Pressure is given by

P=\rho gh

where P=pressure

\rho =Density

h=depth

P=10^3\times 9.8\times 90

P=8.82\times 10^{5}\ N/m^2\approx 8.82\ atm

(b)8.82 atm is much lower than the vapor pressure of water

(c)The fact of applying a negative pressure of 8.74 below the vapor pressure of water

4 0
3 years ago
What is an ellipse? ​
Rzqust [24]

Answer:An ellipse is a closed curve consisting of points whose distances from each of two fixed points (foci) all add up to the same value .

Explanation:

6 0
3 years ago
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