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Varvara68 [4.7K]
3 years ago
13

I need help... anyone know this?

Physics
1 answer:
julsineya [31]3 years ago
7 0

Answer:mp

Explanation:

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the temperature of a 2.0-kg increases by 5*c when 2,000 J of thermal energy are added to the block. What is the specific heat of
nata0808 [166]
To calculate the specific heat capacity of an object or substance, we can use the formula

c = E / m△T

Where
c as the specific heat capacity,
E as the energy applied (assume no heat loss to surroundings),
m as mass and
△T as the energy change.

Now just substitute the numbers given into the equation.

c = 2000 / 2 x 5
c = 2000/ 10
c = 200

Therefore we can conclude that the specific heat capacity of the block is 200 Jkg^-1°C^-1
3 0
3 years ago
A machine lifts a 35 kg object doing 6860 J worth of work. How much power is produced by the machine if it lifts the object in 4
timofeeve [1]

Answer:

Explanation:

We need the power equation for this which is

P = Work/time

We have everything we need to solve this (the mass of the object is extra information):

P = 6860/4

P = 1715W

5 0
3 years ago
Just these 2 question. I have no idea what it's asking..​
Murljashka [212]

Answer: did you get the answers?

Explanation:

5 0
3 years ago
An airplane accelerates from a speed of 88m/s to a speed of 132 m/s during a 15 second time interval. How far did the airplane t
Gelneren [198K]

Answer:

1650\:\mathrm{m}

Explanation:

We can use the following kinematics equations to solve this problem:

v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x.

Using the first one to solve for acceleration:

132=88+a(15),\\15a=44,\\a=\frac{44}{15}=2.9\bar{3}\:\mathrm{m/s^2}.

Now we can use the second equation to solve for the distance travelled by the airplane:

132^2=88^2+2\cdot2.9\bar{3}\cdot \Delta x,\\\Delta x= \frac{9680}{2\cdot2.9\bar{3}},\\\Delta x =\fbox{$ 1650\:\mathrm{m}$}(three significant figures).

6 0
3 years ago
A horizontal disk rotates about a vertical axis through its center. Point P is midway between the center and the rim of the disk
hjlf

Answer:

The answer is explained below.

Explanation:

All the point on the disk has same angular acceleration. Here, the point P is at the midway between the center and the rim of the disk and the point Q is at rim of the disk.

So, the distance of the point Q from the axis is twicee the distance of the point P from the axis.

<em>Rp - R</em>

<em>Rq - 2R</em>

The linear acceleration is

α2 - Rα

So, the linear acceleration of Q is twice as great as the linear acceleration of P.

The speed of the particle when it is in the circular motion depends on the radius of the particle.

In this case, the speed of point Q is twice the speed of point P.

6 0
3 years ago
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