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aleksandrvk [35]
4 years ago
10

A 5.5 kg bowling ball initially at rest is dropped from the top of a 12 m building. It hits the ground 1.75 s later. Find the ne

t external force on the falling ball.
Physics
1 answer:
Thepotemich [5.8K]4 years ago
7 0

We are given that:

m = 5.5 kg

d = 12 m

t = 1.75 s

Using the distance formula, we can get the acceleration of the ball:

d = (1/2) a*t^2

Rearranging:

a = 2d/t^2 = 2*(12 m)/(1.75 s)^2

a = 7.837 m/s^2

 

<span>Using Newton’s 2nd law, force is defines as:</span>

F = ma

F = 5.5 kg (7.837 m/s^2)

<span>F = 43.10 N</span>

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A body of mass m1 = 1.5 kg moving along a directed axis in the positive sense with a velocity
larisa86 [58]

Answer:

3.71 m/s in the negative direction

Explanation:

From collisions in momentum, we can establish the formula required here which is;

m1•u1 + m2•v2 = m1•v1 + m2•v2

Now, we are given;

m1 = 1.5 kg

m2 = 14 kg

u1 = 11 m/s

v1 = -1 m/s (negative due to the negative direction it is approaching)

u2 = -5 m/s (negative due to the negative direction it is moving)

Thus;

(1.5 × 11) + (14 × -5) = (1.5 × -1) + (14 × v2)

This gives;

16.5 - 70 = -1.5 + 14v2

Rearranging, we have;

16.5 + 1.5 - 70 = 14v2

-52 = 14v2

v2 = - 52/14

v2 = 3.71 m/s in the negative direction

8 0
3 years ago
Which of the following is NOT a result of supernova explosions? The neutron core is completely destroyed. Any planets within a f
DanielleElmas [232]

Answer:

 The neutron core is completely destroyed

Explanation:

 A earth - supernova is an explosion resulting to the death of a star that occurs close enough to the earth but this does not completely destroy a star. Supernovae are the most violent explosions in the universe. But they do not explode like a bomb explodes, blowing away every bit of the original bomb. Rather, when a star explodes into a supernova, its core survives. The reason for this is that the explosion is caused by a gravitational rebound effect and not by a chemical reaction. Stars are so large that the gravitational forces holding them together are strong enough to keep the nuclear reactions from blowing them apart. It is the gravitational rebound that blows apart a star in a supernova.

4 0
3 years ago
Vector B has x, y, and z components of 2.4,
Fed [463]

The magnitude of the vector B is 10.9

A vector is a quantity which has magnitude as well as direction and it follows vector laws of addition.

To calculate the magnitude of the vector, we have to put the square of the components of the vector along the axes under the root.

Vector B has components,

x = 2.4

y = 9.8

z = 4.1

Applying the formula,

|B| = √x²+y²+z²

|B| = √(2.4)² + (9.8)² + (4.1)²

|B| = √5.76+96.04+16.81

|B| = √118.61

|B| = 10.9

Talking about the direction the the Vector B, it will be the line joining the origin with the points (2.4,9.8,4.1)

To know more about Vectors, visit,

brainly.com/question/25705666

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3 0
1 year ago
Why would north america will not be able to view the eclipse
Flauer [41]

The North America will not be able to view the eclipse because of its location on the earth caused by the tilting of the earth.

<h3>Effect of tilting of the Earth</h3>

The tilt of the Earth is what causes seasons to occur. These are the seasons in relation to the Northern Hemisphere.

The tilting of the Earth causes the difference in the amount of sun reaching each region of the earth like in the North America continent.

Thus, the North America will not be able to view the eclipse because of its location on the earth caused by the tilting of the earth.

Learn more about eclipse here: brainly.com/question/8643

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4 0
2 years ago
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
3 years ago
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