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Bumek [7]
3 years ago
12

Can someone please help

Physics
1 answer:
ki77a [65]3 years ago
8 0

Answer:

Acceleration of that planet is 30 \frac{m}{s^{2} }.

Given:

initial speed of hammer = 0 \frac{m}{s}

time = 1 s

distance = 15 m

To find:

Acceleration due to gravity = ?

Formula used:

Distance covered by hammer is given by,

s = ut + \frac{1}{2} a t^{2}

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

Solution:

Distance covered by hammer is given by,

s = ut + \frac{1}{2} a t^{2}

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

u = 0

t = 1 s

s = 15 m

a = g

Thus substituting these value in above equation.

15 = 0 + \frac{1}{2} g 1^{2}

g = 15 × 2

g = 30 \frac{m}{s^{2} }

Thus, acceleration of that planet is 30 \frac{m}{s^{2} }.

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telo118 [61]

That is more of a History of English question.

6 0
3 years ago
A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the
sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg

Since the frictional force f_k is equal to f_k=\mu_k N=\mu_k mg, then we have that the acceleration is:

a=\frac{\mu_k mg}{m}=\mu_k g

Now, from the definition of acceleration we get:

a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}

And, as the final velocity is zero because the box gets to a stop, we have:

t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}

Finally, we calculate the displacement:

x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm

This means that the box moves 7.29cm backwards relative to the belt.

4 0
2 years ago
An ice-making machine inside a refrigerator operates in a Carnot cycle. It takes heat from liquid water at 0.0 ∘C and rejects he
Agata [3.3K]

Answer:

2.36 x 10^6 J

Explanation:

Tc = 0°C = 273 K

TH = 22.5°C = 295.5 K

Qc = heat used to melt the ice

mass of ice, m = 85.7 Kg

Latent heat of fusion, L = 3.34 x 10^5 J/kg

Let Energy supplied is E which is equal to the work done

Qc = m x L = 85.7 x 3.34 x 10^5 =  286.24 x 10^5 J

Use the Carnot's equation

\frac{Q_{H}}{Q_{c}}=\frac{T_{H}}{T_{c}}

Q_{H}=286.24\times 10^{5}\times \frac{295.5}{273}

QH = 309.8 x 10^5 J

W = QH - Qc

W = (309.8 - 286.24) x 10^5

W = 23.56 x 10^5 J

W = 2.36 x 10^6 J

Thus, the energy supplied is 2.36 x 10^6 J.

8 0
3 years ago
The grant that considered the foundation of financial aid is the:
navik [9.2K]

Answer:

I think it is the Federal Pell Grant Program.

Explanation:

4 0
2 years ago
Read 2 more answers
A Carnot engine whose high-temperature reservoir is at 464 K has an efficiency of 25.0%. By how much should the temperature of t
nexus9112 [7]

Answer

given,

high temperature reservoir (T_c)= 464 K

efficiency  of reservoir (ε)= 25 %

temperature to decrease = ?

increase in efficiency = 42 %

now, using equation

 \epsilon = 1 - \dfrac{T_C}{T_H}

 0.25 = 1 - \dfrac{T_C}{464}

 T_C= (1 - 0.25) \times 464

 T_C= 0.75 \times 464

      T_C = 348 K

now,

if the efficiency is equal to 42$ = 0.42

 T_C= (1 - 0.42) \times 464

 T_C= 0.58 \times 464

 T_C= 269.12\ K

8 0
3 years ago
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