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liq [111]
3 years ago
9

An object traveling at a constant velocity of 15 m/s is experiencing no net

Physics
1 answer:
Deffense [45]3 years ago
3 0
False
It is experiencing net force
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Mason notices that his boat sinks lower into the water in a freshwater lake than in the ocean. What could explain this?
suter [353]

<u>Answer </u>

D. Salt water is denser than freshwater.

<u>Explanation</u>

A boat is able to float on water because it experiences an upthrust upwards.

The magnitude of the upthrust depends on the density of the liquid.

When the liquid is denser the boat will experience a great upthrust as compared to when in a less dense liquid.

If the boat sinks lower in the freshwater than in salty water, then Salt water is more dense than freshwater.



5 0
3 years ago
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A car goes from point A to point B, five miles away and then returns to point A. The car is going 15 mph.
slavikrds [6]
B will be your answer hope this helped
8 0
3 years ago
A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh
alekssr [168]

(a) 392 N/m

Hook's law states that:

F=k\Delta x (1)

where

F is the force exerted on the spring

k is the spring constant

\Delta x is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

- The stretching of the spring is

\Delta x=15 cm-10 cm=5 cm=0.05 m

Solving eq.(1) for k, we find the spring constant:

k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N

And so, the stretch of the spring is

\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm

And since the initial lenght of the spring is

x_0 = 10 cm

The final length will be

x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm

8 0
4 years ago
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How much power does it take to do 500 J of work in 10 seconds?
Dmitry_Shevchenko [17]
Power = work/time
  
          = 500/10
 
          = 50J/s or 50 watt 


7 0
3 years ago
What are tree things that apply to relative dating
grigory [225]
Rocks, earth aging, fossils
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3 years ago
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