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3 years ago

A ball is dropped off a roof. The ball falls for 4 seconds. Just before the ball hits the ground what is its velocity?

Physics

1 answer:

scZoUnD [109]3 years ago

7 0

Answer:

The velocity is 39.24 m/s

Explanation:

The time of fall t = 4 seconds

The initial velocity u = 0 m/s (since the ball is dropped, meaning it starts from rest)

Final velocity v = ?

Using

v = u + gt

where g is the acceleration due to gravity = 9.81 m/s^2

substituting, we'll have

v = 0 + (9.81 x 4)

v = 39.24 m/s

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A block of 200 g is attached to a light spring with a force constant of 5 N / m and freely in a horizontal plane vibrates. The m

Finger [1]

Answer:

m  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

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2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

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2 2 2.00 0.178 mm  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2 2 2.00 0.178 m

Explanation:

That is a reason

8 0

3 years ago

What speed must a 600 kg car have in order to have the same momentum as a 1200 kg truck traveling at a velocity of 10 m/s to the

yKpoI14uk [10]

Answer:

20 m/s

Explanation:

4 0

2 years ago

A man and his dog are “walking” on flat Street. He is pulling on his stubborn dog with a force of 70 N Directed at a 30° angle f

Delvig [45]

Answer:

X component of force is 60.62 N.

Y component of force is 35 N.

Force of gravity on the dog is 245 N.

Magnitude of normal force is 210 N.

Explanation:

Given:

Force of pull is, $F=70\ N$

Mass of the dog is, $m=25\ kg$

Angle of inclination is, $\theta =30$ °

Acceleration due to gravity is, $g=9.8\ m/s^2$

The free body diagram of the dog is shown below.

The X and Y components of force of pull is given as:

$F_X=F\cos \theta=(70)\cos (30)=60.62\ N\\F_Y=F\sin \theta=(70)\sin(30)=35\ N$

Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.

Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,

$F_g=mg=25\times 9.8=245\ N$

Therefore, the force of gravity on the dog is 245 N.

Now, consider the motion in the vertical direction of the dog. As there is no motion in the vertical direction, the net force along the Y direction is 0. In other words, the total Upward force is equal to the total downward force.

From the free body diagram,

$N+F_Y=mg\\N=mg-F_Y\\N=245-35=210\ N$

Therefore, the normal force acting on the dog is 210 N.

6 0

3 years ago

You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts,

lbvjy [14]

Answer:

The value is $T_c = 12 .1 ^oC$

Explanation:

From the question we are told that

The mass of the ice cube is $m_i = 7.50 *10^{-3} \ kg$

The temperature of the ice cube is $T_i = 0^o C$

The mass of the copper cube is $m_c = 0.540 \ kg$

The final temperature of both substance is $T_f = 0^oC$

Generally form the law of thermal energy conservation,

The heat lost by the copper cube = heat gained by the ice cube

Generally the heat lost by the copper cube is mathematically represented as

$Q = m_c * c_c * [T_c - T_f ]$

The specific heat of copper is $c_c = 385J/kg \cdot ^oC$

Generally the heat gained by the ice cube is mathematically represented as

$Q_1 = m_i * L$

Here L is the latent heat of fusion of the ice with value $L = 3.34 * 10^{5} J/kg$

$Q_1 = 7.50 *10^{-3} * 3.34 * 10^{5}$

=> $Q_1 = 2505 \ J$

$2505 = 0.540 * 385 * [T_c - 0 ]$

=> $T_c = 12 .1 ^oC$

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3 years ago

Question 1 (1 point)

Sonbull [250]

Answer:

Solid objects will deform when adequate loads are applied to them; if the material is elastic, the object will return to its initial shape and size after removal. This is in contrast to plasticity, in which the object fails to do so and instead remains in its deformed state.

Explanation:

4 0

2 years ago

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