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3 years ago

5

Calculate the energy that 1.5 kg of mass contains

1 answer:

Vsevolod [243]3 years ago

4 0

Answer:

$4.5(10)^8$ joules

Explanation:

Mass–energy equivalence is the principle that says that anything that has mass has an equivalent amount of energy. Converse is also true.

The relation between energy and mass is given by $E=mc^2$

Here, E denotes energy, m denotes mass and c denotes speed of light

Mass $(m)=1.5\,\,kg$

Speed of light (c) = 3 × $10^8$ m/s

Therefore,

$E=1.5(3)(10)^8=4.5(10)^8$ joules.

The S.I. unit of energy is joules which is equal to $1\,\,kg\,\,m^2/s^2$

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A 70.0 kg sprinter starts a race with an acceleration of 1.60 m/s^2, What is the net external force (in N) on him? (Enter the ma

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Answer:

External force on him will be 112 N

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Ask Your Teacher Cam Newton of the Carolina Panthers throws a perfect football spiral at 6.9 rev/s. The radius of a pro football

faltersainse [42]

Answer:

$a=159.32\ m/s^2$

Explanation:

It is given that,

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The velocity is given by :

$v=r\omega$

$v=0.085\times 43.35$

v = 3.68 m/s

The centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(3.68)^2}{0.085}$

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3 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

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3 years ago