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3 years ago

List two pieces of evidence related to this picture

Physics

2 answers:

lesya [120]3 years ago

7 0

Crash
Someone upset about the crash

tatyana61 [14]3 years ago

4 0

Answer:

there was a crash you can tell cause of the 2 cars the dog is curious on what happened, you can tell because the dog is looking at the cash

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A sled drops 50 meters in height on a hill. The mass of the rider and sled is 70 kg and the sled is going 10 m/s at the bottom o

navik [9.2K]

Answer:

Efficiency = 10.2 %

Explanation:

Given the following data;

Mass = 70 kg

Height = 50 m

Velocity = 10 m/s

We know that acceleration due to gravity is equal to 9.8 m/s².

To find the efficiency of energy conversion from potential to kinetic;

First of all, we would determine the potential energy;

P.E = mgh

P.E = 70 * 9.8 * 50

P.E = 34300 J

For the kinetic energy;

K.E = ½mv²

K.E = ½ * 70 * 10²

K.E = 35 * 100

K.E = 3500

Therefore, Input energy, I = 34300 J

Output energy, O = 3500 J

Next, we find the efficiency;

Efficiency = O/I * 100

Substituting into the formula, we have;

Efficiency = 3500/34300 * 100

Efficiency = 0.1020 * 100

Efficiency = 10.2 %

4 0

2 years ago

At a playground, a child slides down a slide that makes a 42° angle with the horizontal direction. The coefficient of kinetic fr

kumpel [21]

Answer:

$a = 5.1\ m/s^2$

Explanation:

Given,

The angle of the slide= $42^\circ$

The mass of the child is= m

coefficient of friction = 0.20

when she slides down now apply Newton's law

$ma =mg \sin\theta - f$

$ma = mg\sin \theta -\mu mg \cos \theta$

therefore the acceleration

$a=g \sin\theta -\mu gcos\theta$

$a=g[\sin \theta -\mu \cos \theta]$

$a=9.8\times [\sin 42^\circ -0.2\times \cos 42^\circ]$

$a = 5.1\ m/s^2$

hence, the magnitude of acceleration during her sliding is equal to $a = 5.1\ m/s^2$

4 0

3 years ago

Can you hellp me please

mamaluj [8]

Answer:

i think the answer is B but im not sure

7 0

3 years ago

Read 2 more answers

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with an

Alona [7]

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

$F=\dfrac{kQ^2}{r^2}$ ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

$F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}$ .....(2)

Dividing equation (1) and (2), we get :

$\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}$

Hence, the correct option is (d) i.e. " 4F/9"

7 0

2 years ago

An ideal gas is enclosed in a piston, and 1600 J of work is done on the gas. As this happens, the internal energy of the gas inc

Phantasy [73]

Answer:

- 1100 J heat flows out

Explanation:

dW = - 1600 J (as work is done on the gas)

dU = 500 J

dQ = ?

According to the first law of thermodynamics

dQ = dU + dW

dQ = 500 - 1600

dQ = - 1100 J

As heat is negative so it flows out.

3 0

2 years ago