Series circuit when you had one bulb and battery voltage was at 9 volts, what was current into battery?

A car accelerates uniformly from rest to 20 m/sec in 5.6 sec along a level stretch of road. Ignoring friction, determine the ave

bonufazy [111]

Answer:

(a) $P=33000W$

(b) $P=51000W$

Explanation:

The average power is defined as the amount of work done during a time interval:

$P=\frac{W}{t}(1)$

According to work-energy theorem, the work done is equal to the change in kinetic energy. So, we have:

$W=\Delta K\\W=K_f-K_0\\W=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}\\(2)$

Recall that the weight is given by:

$w=mg\\m=\frac{w}{g}(3)$

The car accelerates uniformly from rest ( $v_0=0$ ). Replacing (3) in (2), we have:

$W=\frac{wv_f^2}{2g}$

(a) Finally, we replace this in (1):

$P=\frac{wv_f^2}{2gt}\\P=\frac{9000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=33000W$

(b)

$P=\frac{14000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=51000W$

3 0

4 years ago

The temperature coefficient of a certain conducting material is 5.74 × 10-3 (°C)-1. (a) At what temperature would the resistance

Alex777 [14]

Answer:

Temperature at which the resistance is twice the resistance at $20^{\circ}C$ is $194.216^{\circ}C$

Solution:

As per the question:

Temperature coefficient, $\alpha = 5.74\times 10^{- 3}^{\circ}C$

Reference temperature, $T_{o} = 20^{\circ}C$

Resistance, $R_{t} = 2R_{o}$

Now, using the formula:

$R_{t} = R_{o}(1 + \alpha \Delta T)$

$2R_{o} = R_{o}(1 + \alpha \times (T _ T_{o}))$

$2 = 1 + 5.74\times 10^{- 3}\times (T - 20^{\circ})$

$\frac{1}{5.74\times 10^{- 3}} = T - 20^{\circ}$

$T = 174.216 + 20 = 194.216^{\circ}$

Yes, this temperature holds for all all the conductors of copper, irrespective of the size and shape of the conductor.

5 0

3 years ago

What is neant by velocity ratio of machine is 3 and efficency is 60%

Kisachek [45]

Answer:

Velocity ratio of a lever is 3 means distance travelled by effort is 3 times the distance travelled by the load. Efficiency of the pulley is 60%means 40% of the energy is lost in the machine due to the friction

3 0

3 years ago

Read 2 more answers

What can you do to mitigate these risks?

andrey2020 [161]

Answer: your question is incomplete, however I want you to know that there are four ways to mitigate a risk, these are:

avoid, accept, reduce/control, or transfer. Whatever the risk in your question is, it will definitely fit into any of the aforemened risk strategies.

7 0

3 years ago

a ball is thrown upward with initial velocity of 20 m/s. (a) how long is the ball in the air? (b) what is the greatest height re

Lena [83]

a) t = 4.08 s

b) Response = 20.39m

c) t = 0.991 and 3.087 seconds.

Initially, the vertical speed was 20 m/s.

(a) how long is the ball in the air?

The ball is tossed upward, rises to its highest point, and then hits the ground again.

Consequently, the end height will be equal to the starting height.

h = h o = Om

Use the equation s= ut + -at.

2764 – An = 14

Put a negative sign for g because the ball is being thrown upwards and g is acting downwards.

Om = t - 0.5 * 9.81 m/s * 20 m/s

0 = 20 *t - 0.5 * 9.81 *t

20 *t=0.5*9.81 *t

20 = 0.5 * 9.81 *t

20 0.5*9.81

Answer (a): t = 4.08 s

(b) what is the greatest height reached by the ball?

Vertical velocity for the ball decreases to zero when it reaches its highest point.

Use the equation 2-u=295

-2ghmas for - u

(20m/s - (Om/s)2) = -2 * 9.81 m/s2 *

02 – 20= -2*9.81 *

-20% = -2 *9.81 *

202 = 2 * 9.81 *

b) RESPONSE: H = 20.39m

(c) when is the ball 15 m above the ground?

The ball will be 15 meters above the earth on two separate occasions.

1. When climbing

2. When descending

Use the equation s= ut + -at.

2,264 – fn = 4

0.5 * 9.81 m/s * 15m = 20 m/s*t

15 = 20 *t - 0.5 * 9.81 *t

15 = 20t – 4.9057

4.905 – 20t + 15 = 0

Calculator-based quadratic equation solution

(c) t = 0.991 and 3.087 seconds.

Hence the answers are,

a) t = 4.08 s

b) RESPONSE: Hmar = 20.39m

c) t = 0.991 and 3.087 seconds.

To learn more about initial velocity, click brainly.com/question/14154244

#SPJ4

4 0

1 year ago