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3 years ago

Give an example in which a small force exerts a large torque. give another example in which a large force exerts a small torque.

Physics

2 answers:

algol [13]3 years ago

5 0

Answer:

Explanation:

Torque is defined as the product of force and the perpendicular distance.

If we take a screw driver of small length then we need to apply a large force to open or tighten the screw.

If a screw driver of long arm, then we need to apply small force to open or tighten the screw.

goldenfox [79]3 years ago

3 0

<span>Since the torque involves the product of force times lever arm, a small force can exert a greater torque than a larger force if the small force has a large enough lever arm.

With a large force exerts a small torque is a gate, hinged in its vertical line (axis). When pushed from a point near to the hinge, a very large amount is needed to open the gate.
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What is the momentum of a 200 kg truck travelling at 20 m/s?

geniusboy [140]

Answer:

p = 4000 kg-m/s

Explanation:

Given that,

The mass of a truck, m = 200 kg

Speed of the truck, v = 20 m/s

We need to find the momentum of the truck. The formula for momentum is given by :

p = mv

so,

$p=200\times 20\\\\p=4000\ kg-m/s$

So, the momentum of the truck is equal to 4000 kg-m/s.

8 0

3 years ago

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.

myrzilka [38]

Answer:

a. The plane speeds up but the cargo does not change speed.

Explanation:

Just to make it clear, the question is as follows from what I understand.

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane. You can neglect air resistance.

Just after the cargo has fallen out:

a. The plane speeds up but the cargo does not change speed.

b. The cargo slows down but the plane does not change speed.

c. Neither the cargo nor the plane change speed.

d. The plane speeds up and the cargo slows down.

e. Both the cargo and the plane speed up.

And we are requested to choose the right answer under the given conditions. We know the glider has no motor, then it must be in free fall movement, then it is experiencing some force that pulls it to the from due to the gravity effect on it, and a force in general is calculated by

F=m*a, m:= mass of the object, a:= acceleration.

Here we are only considering the horizontal effect of the forces, then since the mass is reduced the acceleration must increase to compensate and maintain the equilibrium of the forces, then the glider being lighter can travel faster due to the acceleration. On the other hand by the time the cargo left the glider there was no acceleration and the speed it had at the moment he left the plane continues, then the cargo does not change its speed, then horizontally speaking the answer would be a. The plane speeds up but the cargo does not change speed.

5 0

3 years ago

21. Calculate the acceleration of the bus from point D to E. Show your work.

Marat540 [252]

21) Acceleration from D to E: $1 m/s^2$

22) The acceleration of the bus from D to E is $1 m/s^2$

Explanation:

21)

The acceleration of an object is equal to the rate of change of velocity of the object. Mathematically:

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time elapsed

In this problem, we want to measure the acceleration of the bus from point D to point E. We have:

- Initial velocity at point D: u = 0

- Final velocity at point E: v = 5 m/s

- Time elapsed from D to E: t = 21 - 16 = 5 s

Therefore, the acceleration between D and E is

$a=\frac{5-0}{5}=1 m/s^2$

22) This question is the same as 21), so the result is the same.

Learn more about acceleration:

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4 0

3 years ago

Say you want to make a sling by swinging a mass M of 2.3 kg in a horizontal circle of radius 0.034 m, using a string of length 0

ycow [4]

Answer:

T = 764.41 N

Explanation:

In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:

$F_c=m\frac{v^2}{r}$ (1)

m: mass object = 2.3 kg

r: radius of the circular orbit = 0.034 m

v: tangential speed of the object

However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:

$K=\frac{1}{2}mv^2$

You have the value of the kinetic energy (13.0 J), then, you replace the values of K and m, and solve for v^2:

$v^2=\frac{2K}{m}=\frac{2(13.0J)}{2.3kg}=11.3\frac{m^2}{s^2}$

you replace this value of v in the equation (1). Also, you replace the values of r and m:

$F_c=(2.3kg)(\frac{11.3m^2/s^2}{0.034})=764.41N$

hence, the tension in the string must be T = Fc = 764.41 N

5 0

3 years ago

Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27

densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

$\rho=n\times\dfrac{m}{V_{c}\times N_{A}}$

$n=\dfrac{\rho\timesV_{c}\times N}{m}$ ....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

$\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100$

$\rho=5.98$

We calculate the mass of the alloy

$m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100$

$m=111.15$

Put the value into the equation (I)

$n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}$

$n=1.99\approx 2\ atoms/cell$

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

5 0

3 years ago