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Oksi-84 [34.3K]

2 years ago

11

When the car was stopped by the tree, its change in velocity during the collision was -6 meters/second. This change in velocity

occurred in 2 seconds. What was the acceleration of the car?

1 answer:

jonny [76]2 years ago

8 0

6 meters per second

You will need to add six two times to find the acceleration

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The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be

kap26 [50]

Answer:

$v = 14.86 m/s$

Explanation:

As we know that the force equation at the top is given as

$\frac{mv^2}{R} = ma$

now we know that

$a_c = 1.5 g$

so we have

$\frac{v^2}{R} = 1.5 g$

$v = \sqrt{1.5 Rg}$

so we will have

$v = \sqrt{1.5(15)(9.81)}$

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2 years ago

Which factor indicates the amount of charge on the source charge?

Elina [12.6K]

B. The number of field lines on the source charge.

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A power plant produces 1000 MW to supply a city 40 km away. Current flows from the power plant on a single wire of resistance 0.

nikitadnepr [17]

Answer:

Current = 8696 A

Fraction of power lost = $\dfrac{80}{529}$ = 0.151

Explanation:

Electric power is given by

$P=IV$

where I is the current and V is the voltage.

$I=\dfrac{P}{V}$

Using values from the question,

$I=\dfrac{1000\times10^6 \text{ W}}{115\times10^3\text{ V}} = 8696 \text{ A}$

The power loss is given by

$P_\text{loss} = I^2R$

where R is the resistance of the wire. From the question, the wire has a resistance of $0.050\Omega$ per km. Since resistance is proportional to length, the resistance of the wire is

$R = 0.050\times40 = 2\Omega$

Hence,

$P_\text{loss} = \left(\dfrac{200000}{23}\right)^2\times2$

The fraction lost = $\dfrac{P_\text{loss}}{P}=\left(\dfrac{200000}{23}\right)^2\times2\div (1000\times10^6)=\dfrac{80}{529}=0.151$

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2 years ago

What are ways to improve the design of this experiment? Check all that apply

Serhud [2]

Answer:

B,D,E

Explanation:

I got you

B. Experiment with a wider range of materials.

D.Use a laboratory galvanometer to make precise measurements.

E. Test the strength of the electromagnet by varying the number of wire coils.

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2 years ago

A man wishes to pull a crate 15m across a rough floor by exerting a force of 100 N. The

Darina [25.2K]

Answer:

option (E) is correct.

Explanation:

Work done is defined as the product of force and the distance in the direction of force.

force, f = 100 N

Coefficient of friction, = 0.25

distance = 15 m

So, net force F = f - friction force

F = 100 - 0.25 x m g

Work = (100 - 0.25 mg) x d cosθ

For minimum work, the angle should be maximum.

So, the value of θ is 76°.

thus, option (E) is correct.

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2 years ago