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3 years ago

A handbag weighing 162 N is carried by two students each holding the handle of the bag

Physics

1 answer:

Scrat [10]3 years ago

7 0

Answer:

162 N

Explanation:

Since the handbag weighs 162 N, from Newton's third law of motion, we know that to every action, there is an equal and opposite reaction. Thus, it means the force by each student arms Carrying the bag will be 162 N.

Since both handles of the bag are at same inclined angle of 60°, then the force in each students arm will be equal and thus;

F_each arm = 162 N

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If the period of a wave is 0.05 s, then what is its frequency?

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Frequency = 1/time period = 1/0.05 = 20s^-1.

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A student performs Young’s double-slit experiment with slits that are separated by 0.160 mm. She observes the interference patte

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Explanation:

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3 years ago

The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is $W=1.20 \cdot 10^{-3}J$

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: $W_e=q\Delta V$

where

q is the charge

$\Delta V$ is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

$K_f - K_i = W + W_e = W+q\Delta V$

and since the charge starts from rest, $K_i = 0$ , so the formula becomes

$K_f = W+q\Delta V$

In this problem, we have

$W=1.20 \cdot 10^{-3}J$ is the work done by the external force

$q=-6.70 \mu C=-6.7\cdot 10^{-6}C$ is the charge

$K_f = 4.72\cdot 10^{-4}J$ is the final kinetic energy

Solving the formula for $\Delta V$ , we find

$\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V$

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3 years ago

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Suppose that two objects attract each other with a gravitational force of 32 Newtons. If the distance between the two objects is

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Answer:

64N

Explanation:

32×2= 64N

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3 years ago

The chart show the masses and velocities of two colliding objects that stick together after a collision.

finlep [7]

Answer:

1,500 kg.m/s

Explanation:

First, arrange the information in a table:

Object Mass (kg) Velocity (m/s)

A 200 15

B 150 - 10

After the collision, the two objects are stick together, thus you talk aobut one object and one momentum.

According to the law of convervation of momentum, the momentum after the collision is equal to the momentum before the collision.

Momentum before the collision, P₁:

$P_1=m_{A,1}\times v_{A,1}+m_{B,1}\times v_{B,1}$

$P_1=200kg\times 15m/s+150kg\times (-10m/s)\\\\ P_1=3000kg.m/s-1500kg.m/s=1500kg.m/s$

Momentum after the collision:

As stated, it es equal to the momentum before the collision: 1,500 kg . m/s

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3 years ago