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3 years ago

Pls help me with this question I want the answer ASAP quick

Physics

1 answer:

Sindrei [870]3 years ago

3 0

Answer:

I don't know the answer

Explanation:

I just want the point sorry (\_/)

( - ×)

o o

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Que la tarcer ley de newton

slava [35]

Explanation:

If you write it in English so I can help u if you need it

3 0

3 years ago

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

Katena32 [7]

Answer:

166 W

Explanation:

Power is the rate at which work is done.

$\text{Power} = \dfrac{\text{Work done}}{\text{time}}$

The work done by Jill is the product of the weight of the pail and the height it moves.

The weight is the product of the mass and acceleration of gravity, g. Taking g as 9.81 m/s², the weight is

W = (6.90 kg)(9.81 m/s²) = 67.689 N

Work done = (67.689 N)(27.0 m) = 1827.603 J

Power = (1827.603 J) ÷ (11.0 s) = 166 W

4 0

3 years ago

A 1980-kg car is traveling with a speed of 15.5 m/s. What is the magnitude of the horizontal net force that is required to bring

Dennis_Churaev [7]

Answer: 6067.5 N

Explanation:

Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.

Now, plug this in for work: Work = Force * Distance; so, divide work by distance to get 6067.5 N.

5 0

2 years ago

Compare the two numbers using <, >, =

Effectus [21]

Vas happenin!!

1. =
2. >
3. <
4. <
5. <
6. >

Hope this helps *smiles*

3 0

2 years ago

Lucia kicks a ball on a level playing field with an initial velocity of 11.3 m/s at an angle of 35° above the horizontal. Find:

grandymaker [24]

Explanation:

Given that,

Initial velocity, u = 11.3 m/s

Angle above the horizontal, $\theta=35^{\circ}$

Time of flight :

$t=\dfrac{2u\sin\theta}{g}\\\\t=\dfrac{2\times 11.3\times \sin(35)}{9.8}\\\\t=1.32\ s$

Horizontal distance traveled is given by :

x = ut

x = 11.3 m/s × 1.32 s

x = 14.916 m

Maximum height is given by :

$H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(11.3)^2\times \sin^2(35)}{2\times 9.8}\\\\H=2.14\ m$

Hence, time of flight is 1.32 s, horizontal distance is 14.916 m and maximum height is 2.14 m.

6 0

2 years ago