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lubasha [3.4K]
3 years ago
11

An Apple falls from a tree and one-half second later hits the ground

Physics
1 answer:
docker41 [41]3 years ago
8 0
There isnt enough information to answer the question, the missing variable is "distance from said falling spot and ground"
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If two is company and three is a crowd, what is 4 and 5?
LiRa [457]

4 and 5 is 9

(4 + 5)

4 0
3 years ago
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* CRIMINOLOGY*
Setler [38]

I think the answer is B.

Hope this helps.

5 0
3 years ago
Read 2 more answers
Please help quickly!
Naddik [55]

Answer:

○ D

Explanation:

A resistor is a passive electrical component that uses a circuit element with electric resistance. Resistors are used in electronics for reducing current flow, changing signal rates, dividing voltages, biasing active components, and terminating columns, among other applications.

I am joyous to assist you anytime.

7 0
3 years ago
A 50 N object is on Earth. What is its mass?
faust18 [17]

Answer:   5.10 KG

Explanation:  W = 50 N

                       BY FORMULA

                         W = MG

                          M = W/G

                          M = 50 /9.8

                          M=  5.10 KG

7 0
2 years ago
After being struck by a bowling ball, a 1.3 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.3 kg
GuDViN [60]

Answer:

a) 4.2m/s

b) 5.0m/s

Explanation:

This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.

The problem is also an illustration of elastic collision where there is no loss in kinetic energy.

Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses m_1 and m_2 whose respective velocities before collision are u_1 and u_2;

m_1u_1+m_2u_2=m_1v_1+m_2v_2..............(1)

where v_1 and v_2 are their respective velocities after collision.

Given;

m_1=1.3kg\\u_1=5m/s\\m_2=1.3kg\\u_2=0m/s

Note that u_2=0 because the second mass m_2 was at rest before the collision.

Also, since the two masses are equal, we can say that m_1=m_2=m so that equation (1) is reduced as follows;

mu_1+mu_2=mv_1+mv_2\\\\m(u_1+u_2)=m(v_1+v_2)..............(2)

m cancels out of both sides of equation (2), and we obtain the following;

u_1+u_2=v_1+v_2.............(3)

a) When v_1=0.8m/s, we obtain the following by equation(3)

5+0=0.8+v_2\\hence\\v_2=5-0.8\\v_2=4.2m/s

b) As m_1 stops moving v_1=0, therefore,

5+0=0+v_2\\v_2=5m/s

5 0
3 years ago
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