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Finger [1]
3 years ago
14

A child does 350J of work while pulling a box from the ground up to his tree house with a rope. The tree house is 5.2 m above th

e ground. What is the mass of the box? A) 4.1 kg B) 6.9 kg C) 6.2 kg D) 5.2 kg
Physics
1 answer:
Advocard [28]3 years ago
3 0

I think the answer to that questions is B.

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A projectile is launched at an angle above the
gtnhenbr [62]
The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.

The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.

v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s

Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.

The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.

Combining this together we get:
(1) vx=40m/s and vy=10m/s

7 0
3 years ago
What situation would give you a mechanical advantage? *
const2013 [10]
D. Using a fixed-pulley system
3 0
3 years ago
In this lab, you observed how different factors such as velocity, gradient, and ____ , or amount of water in a stream, affect th
egoroff_w [7]

Answer:

volume and erosion

Explanation:

7 0
3 years ago
Read 2 more answers
A circular curve of radius 150 m is banked at an angle of 15 degrees. A 750-kg car negotiates the curve at 85.0 km/h without ski
Crazy boy [7]

Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.

Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:

Pcos 15°-N=0

Psin15°-f= m*ac

from the first we obtain N, the normal force

N=750Kg*9.8* cos (15°)= 7.1 *10^3 N

Then to calculate the frictional force (f) we can use the second equation

f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r

f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N

6 0
3 years ago
What is the definition of work when net force is parallel to the distance?
goldfiish [28.3K]
Work in general is given by W=F·d where F is the force vector and d is the displacement vector.  The dot symbol is the dot product which is a measure of how parallel two vectors are.  It can be replaced by the cosine of the angle between the two vectors and the vectors replaced by their magnitudes.  If F and d are parallel then the angle is zero and the cosine is unity.  So in this case work can be defined as the product of the magnitudes of the force and distance:
W=Fd
6 0
3 years ago
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