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devlian [24]
1 year ago
8

Any person who opens the door he applies​

Physics
1 answer:
miskamm [114]1 year ago
5 0

Answer:

any person who opens the door he applies pulling force

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1. A2 .7-kg copper block is given an initial speed of 4.0m/s on a rough horizontal surface. Because of friction, the block final
tatyana61 [14]

Answer:

A. Increase in temperature is 0.0176 degree Celsius. b. the remaining energy will be lost.

Explanation:

The mass of copper block = 7kg

Initial speed = 4.0 m/s

Specific heat of copper = 0.385 j/g degree Celcius.

a. The increase in temperature is calculated below:

\text{Change in kinetic energy} = \frac{1}{2}mv^{2} \\=  \frac{1}{2} \times 2.7 \times 4^{2} = 21.6 J \\

85% of energy is converted into internal energy.

mc\DeltaT = 21.6 \times 0.85 \\2.7 \times 385 \times \Delta T = 21.6 \times 0.85 \\\Delta T = 0.0176 degree \ celsius

b. The remaining  15 per cent of kinetic energy will be lost and it will be changed into other forms.

7 0
2 years ago
A block of mass 2.00 kg is initially at rest at x=0 on a slippery horizontal surface for which there is no friction. Starting at
Allisa [31]

Answer:

   x = 1,185 m ,     t = 4/3 s ,  F = - 4 N

Explanation:

For this exercise we use Newton's second law

         F = m a = m dv /dt

        β - α t = m dv / dt

        dv = (β – α t) dt

     

We integrate

        v = β t - ½ α t²

We evaluate between the lower limits v = v₀ for t = 0 and the upper limit v = v for t = t

       v-v₀ = β t - ½ α t²

the farthest point of the body is when v = v₀ = 0

  0 = β t - ½ α t²

  t = 2 β / α

  t = 2 4/6

  t = 4/3 s

Let's find the distance at this time

   v = dx / dt

   dx / dt = v₀ + β t - ½ α t2

   dx = (v₀ + β t - ½ α t2) dt

We integrate

   x = v₀ t + ½ β t - ½ 1/3 α t³

   x = v₀ 4/3 + ½ 4 (4/3)² - 1/6 6 (4/3)³

The body comes out of rest

    x = 3.5556 - 2.37

    x = 1,185 m

The value of force is

    F = β - α t

    F = 4 - 6 4/3

   F = - 4 N

8 0
1 year ago
A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra
skelet666 [1.2K]

Answer:

So the acceleration of the child will be 8.05m/sec^2

Explanation:

We have given angular speed of the child \omega =1.25rad/sec

Radius r = 4.65 m

Angular acceleration \alpha =0.745rad/sec^2

We know that linear velocity is given by v=\omega r=1.25\times 4.65=5.815m/sec

We know that radial acceleration is given by a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2

Tangential acceleration is given by

a_t=\alpha r=0.745\times 4.65=3.464m/sec^

So total acceleration will be a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2

7 0
2 years ago
Sayid made a chart listing data of two colliding objects. A 5-column table titled Collision: Two Objects Stick Together with 2 r
Alborosie

Answer:

6 m/s is the missing final velocity

Explanation:

From the data table we extract that there were two objects (X and Y) that underwent an inelastic collision, moving together after the collision as a new object with mass equal the addition of the two original masses, and a new velocity which is the unknown in the problem).

Object X had a mass of 300 kg, while object Y had a mass of 100 kg.

Object's X initial velocity was positive (let's imagine it on a horizontal axis pointing to the right) of 10 m/s. Object Y had a negative velocity (imagine it as pointing to the left on the horizontal axis) of -6 m/s.

We can solve for the unknown, using conservation of momentum in the collision: Initial total momentum = Final total momentum (where momentum is defined as the product of the mass of the object times its velocity.

In numbers, and calling P_{xi} the initial momentum of object X and P_{yi} the initial momentum of object Y, we can derive the total initial momentum of the system: P_{total}_i=P_{xi}+P_{yi}= 300*10 \frac{kg*m}{s} -100*6\frac{kg*m}{s} =\\=(3000-600 )\frac{kg*m}{s} =2400 \frac{kg*m}{s}

Since in the collision there is conservation of the total momentum, this initial quantity should equal the quantity for the final mometum of the stack together system (that has a total mass of 400 kg):

Final momentum of the system: M * v_f=400kg * v_f

We then set the equality of the momenta (total initial equals final) and proceed to solve the equation for the unknown(final velocity of the system):

2400 \frac{kg*m}{s} =400kg*v_f\\\frac{2400}{400} \frac{m}{s} =v_f\\v_f=6 \frac{m}{s}

7 0
2 years ago
Read 2 more answers
A large solar panel on a spacecraft in Earth orbit produces 1.0 kW of power when the panel is turned toward the sun. What power
Mandarinka [93]

Answer:

e*P_s = 11 W

Explanation:

Given:

- e*P = 1.0 KW

- r_s = 9.5*r_e

- e is the efficiency of the panels

Find:

What power would the solar cell produce if the spacecraft were in orbit around Saturn

Solution:

- We use the relation between the intensity I and distance of light:

                                  I_1 / I_2 = ( r_2 / r_1 ) ^2

- The intensity of sun light at Saturn's orbit can be expressed as:

                                  I_s = I_e * ( r_e / r_s ) ^2

                                  I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2

                                  I_s = 11 W / e*a

- We know that P = I*a, hence we have:

                                  P_s = I_s*a

                                  P_s = 11 W / e

Hence,                       e*P_s = 11 W

3 0
2 years ago
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