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DochEvi [55]
3 years ago
12

how much force ks required.to.keep 780 n box moving at constant velocityu across.the floor if friction force betwern the bix and

the floor is 160 N
Physics
1 answer:
icang [17]3 years ago
8 0

Answer:

Explanation:you must first make a free body diagram of the block to place all the forces that act on the block there. You will have four forces. The normal force, the weight, the horizontal force in x (which is what they ask to calculate) and the friction force. Then you raise the equations and you clear the values. I attach the solution.

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A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00
vivado [14]

Answer:

7.5 m

Explanation:

v = initial speed of the ball = 8 m/s

\theta = angle of launch = 40° deg

Consider the motion along the vertical direction :

v_{oy} = initial velocity along vertical direction = v Sin\theta = 8 Sin40 = 5.14 m/s

a_{y} = acceleration along vertical direction = - 9.8 m/s²

t = time of travel

y = vertical displacement = - 1 m

Using the kinematics equation

y = v_{oy}t + (0.5)a_{y}t^{2}

- 1 = (5.14)t + (0.5)(- 9.8)t^{2

t = 1.22 sec

Consider the motion along the horizontal direction :

v_{ox} = initial velocity along horizontal direction = v Sin\theta = 8 Cos40 = 6.13 m/s

a_{x} = acceleration along vertical direction = 0 m/s²

t = time of travel = 1.22 sec

x = horizontal displacement = ?

Using the kinematics equation

x = v_{ox}t + (0.5)a_{x}t^{2}

x = (6.13)(1.22) + (0.5)(0)(1.22)^{2

x = 7.5 m

4 0
3 years ago
Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o
IrinaVladis [17]

Answer:

The net gravitational force on the mass is 1.27\times 10^{-5}

Explanation:

We have by Newton's law of gravity the force of attraction between masses m_{1},m_{2}

F_{att}=G\frac{m_{1}m_{2}}{r^{2}}

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N

Force of attraction between 435 kg mass and 38 kg mass is

F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N

Thus the net force on mass 38.0 kg is F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}

8 0
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