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3 years ago

Quick Physics Review Questions!!

1 answer:

6 0

I think the correct answer from the choices listed above is the last option. It is only specific heat accounts for mass not heat capacity. Specific heat is specific for a certain amount and has units of <span> J/ kg-K. As you can see, the denominator has kg which is a unit of mass.</span>

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A 62.0 kg sprinter starts a race with an acceleration of 1.44 m/s2. If the sprinter accelerates at that rate for 30 m, and then

Dominik [7]

Answer:

The time taken for the race is 17.20 s.

Explanation:

It is given in the problem that a 62.0 kg sprinter starts a race with an acceleration of 1.44 meter per second square.The initial speed of the sprinter is zero as it starts from the rest.

Calculate the final speed of the sprinter.

The expression for the equation of the motion is as follows;

$v^{2}-u^{2}= 2as$

Here, u is the initial speed, v is the final speed, a is the acceleration and s is the distance.

Put u= 0, s=30 m and $a= 1.44 ms^{-2}$ .

$v^{2}-(0)^{2}= 2(1.44)(30)$

$v= 9.3 m^{-1}$

Calculate time taken to cover 30 m distance.

The expression for the equation of motion is as follows;

$s=ut+\frac{(1)}{2}at^2$

Put u= 0, s=30 m and $a= 1.44 ms^{-2}$ .

$30=(0)t+\frac{(1)}{2}(1.44)t^2$

t=6.45 s

Calculate the time taken to complete his race.

T= t+t'

Here, t is the time taken to cover 30 m distance and t' is the time taken to cover 100 m distance.

$T=6.45+\frac{s'}{v}$

Put s= 30 m, $v= 9.3 m^{-1}$ and s'= 100 m.

$T=6.45+\frac{(100)}{9.3}$

T= 17.20 s

Therefore, the time taken for the race is 17.20 s.

6 0

3 years ago

One Pro for supplements is the cost.<br> True<br> Or False

Nat2105 [25]

Answer:

False

Explanation:

8 0

3 years ago

You push a 85 kg shopping cart from rest with a net force of 250 n for 5 seconds,at which point it flies off a cliff that is 100

Vikki [24]

m = mass of the cart = 85 kg

F = net force on the cart = 250 N

a = acceleration of the cart

acceleration of the cart is given as

a = F/m

a = 250/85

a = 2.94 m/s²

t = time for which the force is applied = 5 sec

v₀ = initial velocity of the cart = 0 m/s

v = final velocity of the cart just before it flies off the cliff = ?

using the equation

v = v₀ + a t

inserting the values

v = 0 + (2.94) (5)

v = 14.7 m/s

consider the motion of cart after it flies off the cliff in vertical direction :

v' = initial velocity in vertical direction = 0 m/s

a' = acceleration in vertical direction = g = acceleration due to gravity = 9.8 m/s²

t' = time taken for the cart to land = ?

Y' = vertical displacement of the cart = height of cliff = 100 m

using the kinematics equation

Y' = v' t' + (0.5) a' t'²

100 = (0) t' + (0.5) (9.8) t'²

t' = 4.52 sec

consider the motion of cart along the horizontal direction after it flies off the cliff

X = distance traveled from the base of cliff = ?

t' = time of travel = 4.52 sec

v = velocity along the horizontal direction = 14.7 m/s

distance traveled from the base of cliff is given as

X = v t'

X = 14.7 x 4.52

X = 66.4 m

3 0

3 years ago

Does a comets tail always trail along behind it in its orbit?

Marat540 [252]

No, it only does when entering an atmosphere

4 0

3 years ago

Two balls are at the same height and released at the same time. One ball is dropped and hits the ground 5s later. The 2nd ball i

ASHA 777 [7]

Answer:

The second ball hits at the same time.

Distance travelled is 65 meters

Explanation:

5 0

1 year ago