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vichka [17]
3 years ago
11

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

Physics
1 answer:
Vinil7 [7]3 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 force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching i
Sunny_sXe [5.5K]

Answer:

114.44 J

Explanation:

From Hook's Law,

F = ke................. Equation 1

Where F = Force required to stretch the spring, k = spring constant, e = extension.

make k the subject of the equation

k = F/e.............. Equation 2

Given: F = 10 lb = (10×4.45) N = 44.5 N, e = 4 in = (4×0.254) = 1.016 m.

Substitute into equation 2

k = 44.5/1.016

k = 43.799 N/m

Work done in stretching the 9 in beyond its natural length

W = 1/2ke²................. Equation 3

Given: e = 9 in = (9×0.254) = 2.286 m, k = 43.799 N/m

Substitute into equation 3

W = 1/2×43.799×2.286²

W = 114.44 J

3 0
3 years ago
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Which statement is true for electric field lines?
KiRa [710]
The best answer is D. field lines should always be crossing each other.
8 0
4 years ago
Read 2 more answers
State keplers law........​
S_A_V [24]

Answer:

There are actually three, Kepler's laws that is, of planetary motion: 1) every planet's orbit is an ellipse with the Sun at a focus; 2) a line joining the Sun and a planet sweeps out equal areas in equal times; and 3) the square of a planet's orbital period is proportional to the cube of the semi-major axis of its

4 0
3 years ago
A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 63.0°-angle with the horizontal.
crimeas [40]

Answer:

Explanation:

Given:

length of ladder r_L = 14m

weight of ladder F_L = 490N

position of firefighter r_F = 3.8m

weight of firefighter F_F = 820N

angle of ladder \alpha = 63

Unknown:

force of the wall on the ladder F_W

force of friction on base of ladder F_R

normal force on base of ladder F_N

From the free body diagram of the sketch you get 3 equations:

F_x = ma_x = F_W - F_R = 0\\ F_y = ma_y = F_N - F_F - F_L = 0\\ \tau _P = \overrightarrow{r} \times \overrightarrow{F} = r_FF_Fcos\alpha + \frac{1}{2}r_LF_Lcos\alpha - r_LF_Wsin\alpha = 0

Solving the equations gives:

F_W = F_R\\ F_N = F_F + F_L\\ F_W = \frac{r_FF_F + 0.5r_LF_L}{r_L tan\alpha}

a)

F_R = 238N\\ F_N = 1310N

b)

F_R = \mu F_N\\ \mu = \frac{F_R}{F_N} \\ \mu = 0.3

c) Using the result from b and solving for r_F

\\ \mu = 0.15\\ F_R = \mu F_N\\ r_F = 2.4m

4 0
4 years ago
On what factors does critical velocity depend on
ASHA 777 [7]

Explanation:

The critical velocity is that velocity of liquid flow, up to which its flow is streamlined (laminar)& above which its flow becomes turbulent. It's denoted by Vc & it depends upon: Coefficient of viscosity of liquid (η) Density of liquid. Radius of the tube.

6 0
3 years ago
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