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2 years ago

The electric motor in the car is powered by a battery.

Physics

1 answer:

Vinil7 [7]2 years ago

6 0

Answer:

I = 30 A.

Explanation:

Given that,

The voltage of the battery, V = 230 V

Power used to charge the battery, P = 6.9 kW

We need to find the current used to charge the battery. The formula for the power is given by :

P = VI

Where

I is current

So,

So, the required current is 30 A.

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A basketball player throws a chall -1 kg up with an initial speed of his hand at shoulder height = 2.15 m Le gravitational poten

Talja [164]

Complete Question:

A basketball player tosses a basketball m=1kg straight up with an initial speed of v=7.5 m/s. He releases the ball at shoulder height h= 2.15m. Let gravitational potential energy be zero at ground level

a) Give the total mechanical energy of the ball E in terms of maximum height hn it reaches, the mass m, and the gravitational acceleration g.

b) What is the height, hn in meters?

Answer:

a) Energy = mghₙ

b) Height, hₙ = 5.02 m

Explanation:

a) Total energy in terms of maximum height

Let maximum height be hₙ

At maximum height, velocity, V=0

Total mechanical energy , E = mgh + 1/2 mV^2

Since V=0 at maximum height, the total energy in terms of maximum height becomes

Energy = mghₙ

b) Height, hₙ in meters

mghₙ = mgh + 1/2 mV^2

mghₙ = m(gh + 1/2 V^2)

Divide both sides by mg

hₙ = h + 0.5 (V^2)/g

h = 2.15m

g = 9.8 m/s^2

V = 7.5 m/s

hₙ = 2.15 + 0.5(7.5^2)/9.8

hₙ = 2.15 + 2.87

hₙ = 5.02 m

6 0

3 years ago

Plz i need help for the 5 problems. plz show the work!!!

Artemon [7]

Answer:

1. 3 $m/s^{2}$

2. 1.5 $m/s^{2}$

3. 3 seconds

4. 0 $m/s^{2}$

5. 2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

$a=\frac {v-u}{t}$

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = $\frac {30-0}{10}=3 m/s^{2}$

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

$a=\frac {v-u}{t}$

Substituting u=10m/s, v=22m/s and t=8 seconds

a = $\frac {22-10}{8}=1.5 m/s^{2}$

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

$t=\frac {v-u}{a}$

Substituting u=0m/s since at rest, v=15m/s and a=5 $\frac {m}{s^{2}}$

= $\frac {15-0}{5}=3s$

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

$a=\frac {v-u}{t}$

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = $\frac {20-20}{10}=0 m/s^{2}$

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

$t=\frac {v-u}{a}$

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 $\frac {m}{s^{2}}$

= $\frac {0-9}{-4.1}=2.2s$

8 0

3 years ago

What is the significance of the nose end marking on a rocket or missile?

Gemiola [76]

If you're referring to the different colors that usually occur at the tip of missles, rockets and some other aircraft, it either a) signifies the end of a particular plate of metal, fabricated specifically to be for the nose. Sometimes these can even be a different alloy or metal all together. or b) this shows where the curved surface begins, so in the case of damage or imperfections due to wear, they can be repaired and measured more easily. The shape of the nose is extremely important for smooth flight, and a dent or bump formed on it can make the aircraft unstable. If you can measure from where the curve starts by the difference in color, it makes repairing or re-fabricating the part much easier. Many of these curves aren't as simple as they appear.

5 0

3 years ago

Consider three identical electric bulbs of power P. Two of bulbs are connected in series and the third one is connected in paral

Tamiku [17]

Answer: 3P/2

Explanation: Let the resistance of the bulbs be R.

now lets consider a Voltage V is supplied to the parallel circuit such that

$P=VI=V^2/R$