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2 years ago

An object with a mass of 78kg is lifted through a height of 6 meters. How much work is done?

1 answer:

3 0

Work = force x distance.

force = mass x acceleration

work = mass x acceleration x diastance

use acceleration of gravity in this problem

W (J) = m (kg) x a (m/s/s) x d (m)
W = 78 x 9.8 x 6
W = 4586.4

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Spiderman, whose mass is 74.0 kg, is dangling on the free end of a 11.0-m-long rope, the other end of which is fixed to a tree l

Anestetic [448]

Answer:

W = -1844.513 J

Explanation:

GIVEN DATA:

mass of spider man is m 74 kg

vertical displacement if spider is 11 m

final displacement = 11 cos 60.6 = - 6.753 m

change in displacement is = -6.753 - (-11) = 4.25 m

gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N

work done by gravity is $W = F \delta r cos\theta$

$W = 725.2 \times 4.25 \times cos 180$

where 180 is the angle between spiderman weight and displacement

W = -1844.513 J

7 0

2 years ago

What type of charge does an electron have?

GarryVolchara [31]

An electron has a negative charge. Hope this helps.

4 0

2 years ago

Read 2 more answers

An object has traveled 20 meters in 40 seconds what is its average speed

Mice21 [21]

Hello,

To solve we need to know the formula for speed

The formula is D/T=S (Distance of time=speed)

Now all we have to do is plug in the numbers.

20/40= 1/2 or 0.5

SO the speed is 0.5 m/s

Have a great day!

6 0

2 years ago

Helppppppp pleaseeee

SVETLANKA909090 [29]

Answer: (Sorry, but I don't know how to calculate mass)

1. 15 N

2. 0.4921 $\frac{ft}{s^2}$ (feet per second squared)

4. 150 N

5. 8.202 feet per second squared

3 0

1 year ago

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

6 0

2 years ago