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3 years ago

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When landing after a spectacular somersault, a 35.0 kg gymnast decelerates by pushing straight down on the mat. calculate the fo

rce she must exert if her deceleration is 9.00 times the acceleration of gravity?

1 answer:

Sliva [168]3 years ago

6 0

The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.

Now from Newton`s first law, the net force on gymnast,

$F_{net} =F-W=ma$

Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast ( $9\times g$ acceleration due to gravity)

Therefore,

$F= ma+W$ OR $F=ma+mg=m(g+a)$

Given $m = 30 kg$ and $a=9\times g=9\times 9.8 m/s^{2} =88.2 m/s^{2}$

Substituting these values in above formula and calculate the force exerted by the gymnast,

$F=(40 kg) (88.2 m/s^{2} +9.8 m/s^{2} )$

$F=3.537\times10^{3}N$

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One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction

goblinko [34]

Answer:

The magnitud of the force is 124.8N.

Explanation:

First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:

$0=F-FrictionAB$

$F=FrictionAB=Nab*$ μs

and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:

$0=Nab-W$

$Nab=W=mg$

replacing this we have:

$F=$ μs $*Nab=$ μs* $mg=41.6N$

and μs $=41.6N/(mg)$

now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.

the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.

now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.

again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:

$F-Friction1(ground)-Friction2(AB)=0$

This is the new external F force that we are looking for:

$F=Friction1(ground)+Friction2(AB)$

we know Friction2(AB) because we found that in the previous block so:

$F=Friction1(ground)+mg$ *μs

for the other friction we have to use the equation:

$Friction(ground)=N(ground)$ *μs

from y axis we have:

$N(ground)-w-Normal(AB)=0$

$N(ground)=w+Normal(AB)$

we found the value of Normal(AB) with the previous block so:

$N(ground)=mg+mg=2mg$

and:

$Friction(ground)=2mg$ *μs

$F=Friction(ground)+mg$ *μs

$F=2mg$ *μs+μs* $mg=3mg$ *μs

and since: μs* $mg=41.6N$

the new F force would be:

$F=3mg$ *μs $=41.6*3=124.8N$

4 0

3 years ago

A 12.0kg microwave oven is pushed 14.0m up the sloping surface of a loading ramp inclined at an angle 37 degrees above the horiz

Semenov [28]

Answer:

Answered

Explanation:

a) What is the work done on the oven by the force F?

W = F * x

W = 120 N * (14.0 cos(37))

<<<< (x component)

W = 1341.71

b) $F_f=\mu_k N$

$F_f=0.25\times12\times9.8$

= 29.4 N

$W_f= F_f\times x$

$W_f= 29.0\times 14 cos(37)$

W_f= 328.72 J = 329 J

c) increase in the internal energy

U_2 = mgh

= 12*9.81*14sin(37)

= 991 J

d) the increase in oven's kinetic energy

U_1 + K_1 + W_other = U_2 + K_2

0 + 0 + (W_F - W_f ) = U_2 + K_2

1341.71 J - 329 J - 991 J = K_2

K_2 = 21.71 J

e) F - F_f = ma

(120N - 29.4N ) / 12.0kg = a

a = 7.55m/s^2

vf^2 = v0^2 + 2ax

vf^2 = 2(7.55m/s)(14.0m)

V_f = 14.5396m/s

K = 1/2(mv^2)

K = 1/2(12.0kg)(14.5396m/s)

K = 87.238J

4 0

2 years ago

Give two ways of reversing the direction of the forces on the coil in the electric motor?

nikklg [1K]

Answer:

Interchanging the poles of the magnet

Reversing the direction of the applied current

Explanation:

The working of the electric motor is associated with Fleming's left-hand rule.
It states that if a current-carrying conductor is placed inside a magnetic field, it experiences a force in the direction perpendicular to the direction of the electric current and magnetic field.
These three physical quantities are placed in a mutually perpendicular direction.
So, in order to reverse the direction of force, you have to either change the direction of the current or magnetic field.

4 0

3 years ago

An Amtrak going 250m/s comes to a stop in 12s. What is the<br> acceleration?

astraxan [27]

Answer:

$a=\frac{v-u}{t} \\a = \frac{0-250}{12} = -20.83 m/s$

Explanation:

you mean deceleration right ? because the acceleration is 250m/s

7 0

3 years ago

An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu

Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

$\alpha = \frac{\omega_f - \omega_i}{t}$

where we have

$\alpha = -80.0 rad/s^2$ is the angular acceleration (negative since the rotor is slowing down)

$\omega_f$ is the final angular speed

$\omega_i = 2000 rad/s$ is the initial angular speed

t = 10.0 s is the time interval

Solving for $\omega_f$ , we find the final angular speed after 10.0 s:

$\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s$

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

$\alpha = \frac{\omega_f - \omega_i}{t}$

If we re-arrange it for t, we get:

$t = \frac{\omega_f - \omega_i}{\alpha}$

where here we have

$\omega_i = 2000 rad/s$ is the initial angular speed

$\omega_f=0$ is the final angular speed

$\alpha = -80.0 rad/s^2$ is the angular acceleration

Solving the equation,

$t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s$

6 0

3 years ago