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3 years ago

A car accelerates from rest (zero speed) up to a speed of 30m/s in 12

Physics

1 answer:

il63 [147K]3 years ago

7 0

Answer:

Explanation:

v=u+at

30=0+a×12

30=12a

a= $\frac{30}{12}$

a=2.5ms⁻²

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A pail in a water well is hoisted by means of a frictionless winch, which consists of a spool and a hand crank. When Jill turns

Sveta_85 [38]

Answer:

182.28 W

Explanation:

Here ,

m = 7.30 Kg

distance , d= 28.0 m

time , t = 11.0 s

average power supplied = change in potential energy/time

average power supplied = m×g×d/time

average power supplied = 7.30×9.81×28/11

average power supplied = 182.28 W

the average power supplied is 182.28 W

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2 years ago

tom and ted are sitting on seprate chairs that have wheels. tom pushes ted and, in turn, he starts moving too. in which directio

IceJOKER [234]

In the direction opposite to his push.

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3 years ago

Short Answer

STatiana [176]

15 degrees because a glass of water won't do anything to a bath tub of 15 degree water

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3 years ago

A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizo

Alla [95]

Answer:

$v_{ox}= 19.6\ m/s$

Explanation:

Data provided in the question:

Height above the ground, H= 5.0m

Range of the ball, R= 20 m

Initial horizontal velocity = $v_{ox}$

Initial vertical velocity= $v_{oy}$ (Since ball was thrown horizontally only)

Acceleration acting horizontally, $a_x$ = 0 m/s² [ Since no acceleration acts horizontally) ]

Vertical Acceleration, $a_y$ = 9.8 m/s² (Since only gravity acts on it)

Let 't' be the time taken to reach ground

Therefore, using equations of motion, we have

$H= v_{oy}t+\frac{1}{2}a_yt^2$

$5= (0)t+\frac{1}{2}(9.8)t^2$

$t= \frac{10}{9.8}=1.02 s$

Then using Equations of motion for horizontal motion,

$R= v_{ox}t+\frac{1}{2}a_xt^2$

$20= v_{ox}(1.02)+\frac{1}{2}(0)(1.02)^2$

$v_{ox}= 19.6\ m/s$

4 0

3 years ago

Say I have a series circuit with 20v and four 65 ohm resistors, what is the current in each resistor?

Komok [63]

Data:

$E = 20 V$
$R_{1} = 65\Omega$
$R_{2} = 65\Omega$
$R_{3} = 65\Omega$
$R_{4} = 65\Omega$

Now that we have all the values we need properly identified, simply calculate the equivalent total resistance of the circuit and the intensity of the total electric current using the Ohm's Law:

 $R_{T} = R_{1} + R_{2} + R_{3} + R_{4}$
$R_{T} = 65 + 65 + 65+ 65$
$R_{T} = 260\Omega$

Like this:

$I_{T} = \frac{E}{ R_{T} }$

$I_{T} = \frac{20}{ 260 }$
$I_{T} = 0,076923076...$

$\boxed{\boxed{I_{T} \approx 0,07A}}$
Answer:
The intensity of the total electric current
 $\boxed{\boxed{I_{T} \approx 0,07A}}$

P.S:. Since the association is in series, the current of 0.07A is the same for all resistors.

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3 years ago