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2 years ago

Please help me I have to send for my teacher

Physics

1 answer:

expeople1 [14]2 years ago

8 0

Answer:

for the fill in the blanks

1- static

2-kinetic

3-coeffiecient

4-opposite to

5-sin theta

6-cos theta

im not sure however what to do with the top part or if its even part of what you need help with

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A student pushed a box 32.0 m across a smooth, horizontal floor using a constant force of 124 N. If the force was applied for 8.

Brilliant_brown [7]

The power developed is 500 W ( to the nearest Watt)

Power(P) is the rate at which work is done. Work done (W) is the product of the force applied on the object and the displacement (s) made by the point of application of the force.

$P = \frac{W}{t}$

$W= F*s$

Therefore,

$P=\frac{F*s}{t}$

Substitute the given values of force , displacement and time

$F = 124 N,s = 32.0 m,t = 8.0 s$

$P =\frac{W*s}{t} =\frac{124N*22.0s}{8.0s} =496 W$

Thus the Power can be rounded off to the nearest value of 500 W

3 0

3 years ago

Read 2 more answers

What is the molecular geometry or of the molecular molecule

NeTakaya

Molecular Geometry is the three-dimensional structure or arrangement of atoms in a molecule.

5 0

3 years ago

Suppose you push a hockey puck of mass m across frictionless ice for a time \Delta t, starting from rest, giving thepuck speed v

bazaltina [42]

Answer:

1. $t_2 = 2t_1$

2. $t_2 = t_1\sqrt{2}$

Explanation:

1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.

In Newton's 2nd law: F = ma

where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

$a_1 = 2a_2$

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

$t = v / a_1$

Now that acceleration is halved:

$t = \frac{v}{2a_2}$

$\frac{v}{a_2} = 2t$

You would need to push for twice amount of time $t_2 = 2t_1$

2. The distance traveled by the puck is as the following equation:

$d = at^2$

So if the acceleration is halved while maintaining the same d:

$\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}$

As $d_1 = d_2$ , then $d_1/d_2 = 1$ . Also $a_1 = 2a_2$

$1 = \frac{2a_2t_1^2}{a_2t_2^2}$

$t_2^2 = 2t_1^2$

$t_2 = t_1\sqrt{2}\approx 1.14t_1$

So t increased by 1.14

7 0

3 years ago

A 170 kg astronaut (including space suit) acquires a speed of 2.25 m/s by pushing off with his legs from a 2600 kg space capsule

saw5 [17]

Explanation:

Mass of the astronaut, m₁ = 170 kg

Speed of astronaut, v₁ = 2.25 m/s

mass of space capsule, m₂ = 2600 kg

Let v₂ is the speed of the space capsule. It can be calculated using the conservation of momentum as :

initial momentum = final momentum

Since, initial momentum is zero. So,

$m_1v_1+m_2v_2=0$

$170\ kg\times 2.25\ m/s+2600\ kg\times v_2=0$

$v_2=-0.17\ m/s$

So, the change in speed of the space capsule is 0.17 m/s. Hence, this is the required solution.

8 0

3 years ago

A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The ang

olga55 [171]

Answer:

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child: $s_c = \omega_ct = 0.233t$