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omeli [17]
3 years ago
9

How do motion and Newton's laws apply to your everyday life? (all the laws)

Physics
1 answer:
Nataly [62]3 years ago
4 0

For me personally, Newton's third law of motion are ubiquitous in everyday life. For example, when you jump, your legs apply a force to the ground, and the ground applies and equal and opposite reaction force that propels you into the air.

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In physics what does 7.56 × 5.746 equal ?
lapo4ka [179]

Answer:

43.43

Explanation:

5.746 x 7.56 = 43.43976

As the least number of desimal is two so our awnser should contain two digits after the decimal point.

Ans: 43.43.

7 0
3 years ago
Sandy is on a road trip. She leaves at 8:00 AM. It takes her 2 hours to drive 200 kilometers. She stops at a rest stop for half
lutik1710 [3]
The average velocity of Sandy is given by the total distance covered S divided by the total time taken t:
v= \frac{S}{t}

The total distance covered is
S=200 km+0+100 km=300 km
while the total time taken is 2 hours + half an hour (for the rest) + 1 hour and half, so
t=2h+0.5h+1.5 h=4 h
Therefore, the average velocity is 
v= \frac{S}{t}= \frac{300 km}{4 h}=75 km/h
0 0
3 years ago
An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the m
Aneli [31]

Answer:

\displaystyle y_m=3.65m

Explanation:

<u>Motion in The Plane</u>

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of v_o and \theta\\ as the initial speed and angle, then we have

\displaystyle v_x=v_o\ cos\theta

\displaystyle v_y=v_o\ sin\theta-gt

\displaystyle x=v_o\ cos\theta\ t

\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}

If we want to know the maximum height reached by the object, we find the value of t when v_y becomes zero, because the object stops going up and starts going down

\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt

Solving for t

\displaystyle t=\frac{v_o\ sin\theta }{g}

Then we replace that value into y, to find the maximum height

\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2

Operating and simplifying

\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}

We have

\displaystyle v_o=20\ m/s,\ \theta=25^o

The maximum height is

\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}

\displaystyle y_m=3.65m

7 0
3 years ago
A wave traveling in water has a frequency of 500.0 Hz and a wavelength of 3.00 m. What is the speed of the wave?
Sergeu [11.5K]

Answer:

1500 m/s

Explanation:

Recall that for a wave,

Speed = frequency x wavelength

here we are given frequency = 500 Hz and wavelength = 3m

simply substitute into above equation

Speed = 500 Hz x 3m

= 1500 m/s

6 0
3 years ago
Read through the scenarios below and calculate the predicted change in kinetic energy of the object compared to a 100 kg ball tr
sashaice [31]

Okay, first off, the formula for Kinetic Energy is:

<em>KE = 1/2(m)(v)^2</em>

<em>m = mass</em>

<em>v = velcoity (m/s)</em>


Using this formula, we can then calculate the kinetic energy in each scenario:

1) KE = 1/2(100)(5)^2 = 1,250 J

2) KE = 1/2(1000)(5)^2 = 12,500 J

3) KE = 1/2(10)(5)^2 = 125 J

4) KE = 1/2(100)(5)^2 = 1,250 J

4 0
3 years ago
Read 2 more answers
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