Ask question

Ask question

All categories

3 years ago

11

because the snow suddenly gets too slushy, you decide to carry your 100-newton sled the rest of the way home. How much workdo yo

u you do when you pick up your sled, lifting it 0.5meter upward? How much work do you do to carry the sled if your house is 800 meters away?

1 answer:

quester [9]3 years ago

3 0

When you lift it the work is 50 joules. When heading home the work is 80000joules

You might be interested in

Light travels in a straight line at a constant speed of 300 000 km/s what is the lights acceleration

exis [7]

Answer:

it's acceleration is 0

Explanation:

since it is travelling at a constant speed it is not accelerating so its acceleration is 0

3 0

3 years ago

The diagram shows forces acting on a boat.

Greeley [361]

A

I hope this helps!!:)

7 0

3 years ago

According to the barometer in the attachment below, which statement is true?

mixer [17]

Your answer is A

Pls mark me brainiest and I sure hope this helps you

8 0

2 years ago

Read 2 more answers

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind

Marat540 [252]

Answer:

Part a)

$F_v = 4.28 N$

Part B)

$L = 1.02 m$

Part C)

$v = 1.25 m/s$

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

$Tcos\theta = mg$

$T sin\theta = F_v$

$\frac{F_v}{mg} = tan\theta$

$F_v = mg tan\theta$

$F_v = 1.2\times 9.81 (tan20)$

$F_v = 4.28 N$

Part B)

Here we can use energy theorem to find the distance that it will move

$-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2$

$(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2$

$(-3.5 + 2.54)L = - 0.98$

$L = 1.02 m$

Part C)

At terminal speed condition we know that

$F_v = mg$

$bv^2 = mg$

$2.5 v^2 = 3.9$

$v = 1.25 m/s$

7 0

3 years ago

Mathphys this is the last time i ever ask for help PLEASEHELP ME

Ivenika [448]

Explanation:

Heat flow = conductivity × area × change in temperature / thickness

q' = kAΔT/t

13.3 W = k (1.56 m²) (7.8°C) / (0.0234 m)

k = 0.0256 W/m/°C

Heat lost by water = heat gained by ice

-mCΔT = mL + mCΔT

-(1000 g) (1 cal/g/°C) (12°C − 37°C) = m (79.7 cal/g) + m (1 cal/g/°C) (12°C − 0°C)

25,000 cal = (91.7 cal/g) m

m = 272.6 g

8 0

3 years ago