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maw [93]
3 years ago
15

A nurse applies a horizontal force of 147 N on a bed that has a mass of 152 kg,The magnitude of the normal force acting on the b

ed is
.967 N
5.00 N
1440 N
1490 N
Physics
1 answer:
oee [108]3 years ago
3 0

Answer:

1490 N

Explanation:

The force applied by the nurse is horizontal, so there's no vertical component.

Therefore, the normal force is simply equal to the weight of the bed.

N = mg

N = (152 kg) (9.8 m/s²)

N = 1490 N

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7 0
3 years ago
A:10i - 2j -4k and B: i +7j - k. Determine |A-B| ​
maksim [4K]

<em>A</em> - <em>B</em> = (10<em>i</em> - 2<em>j</em> - 4<em>k</em>) - (<em>i</em> + 7<em>j</em> - <em>k</em>)

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3 years ago
What is the name of the element with 83 protons and 80 electrons?
dmitriy555 [2]
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4 years ago
A general definition of media is "methods for communicating information." Please select the best answer from the choices provide
igor_vitrenko [27]

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6 0
4 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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