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kobusy [5.1K]
3 years ago
10

What is the speed of a bobsled whose distance-time graph indicates that it traveled 107m in 27s?

Physics
1 answer:
Snowcat [4.5K]3 years ago
8 0
V=D/T
V=107 m / 27 s - divide distance by time
V= 4 m/s - because when we round 3.9629 in the nearest 10th we will get 4. 
Hope this helps

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In terms of newton first law of motion why is it important to wear a seat belt
lilavasa [31]

Answer:

Seatbelts stop you

Explanation:

Any passengers in the car will also be decelerated to rest if they are strapped to the car by seat belts.

6 0
4 years ago
How much heat is removed from 60 grams of steam at 100 °C to change it to 60 grams
Harrizon [31]

Answer:

45200J

Explanation:

Given parameters:

Heat of vaporization of water  = 2260J/g

Mass of steam = 20g

Temperature = 100°C

Unknown:

Energy released during the condensation  = ?

Solution:

This change is a phase change and there is no change in temperature

To find the amount of heat released;

         H  = mL

m is the mass

L is the latent heat of vaporization

Insert the parameters and solve;

         H  = 20g x 2260J/g

          H = 45200J

4 0
2 years ago
according to newton's second law of motion of the net force acting on the object increases while the mass of the object remains
Licemer1 [7]

Answer:

The Acceleration will increase

Explanation:

Newton's Second Law of motion: It states that the rate of change of momentum is directly proportional to the applied force and takes places along the direction of the force.

It can be expressed mathematically as,

F ∝ m(v-u)/t

Where (v-u)/t = a

F  = kma.

F = force, m = mass of the body, a = acceleration, k = constant of proportionality which tend to unity for a unit force, a unit mass, and a unit acceleration.

Therefore,

F = ma.

From the equation above,

If the net force acting on a body increase, while the mass of the body remains constant, the acceleration will also increase.

4 0
3 years ago
A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh
DiKsa [7]

Answer:(a)360N,(b)171N,(c)2.702m

Explanation:

(a)Maximum Friction Force =\mu \left ( N\right )=0.4\times \left ( 740+160\right )

=360 N

cos\theta =\frac{3}{5}

sin\theta =\frac{4}{5}

(b)Moment about Ground Point

740\times 1\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta

N_1tan\theta =1140

N_1=171 N

N_1=f=171 N

(c)

740\times x\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta

Here maximum friction force can be 360 N

Therefore N_1=360 N

Where x is the maximum distance moved by man along the ladder

360\times 5\times \frac{4}{3}=740x+160\times 2.5

740x=2000

x=2.702m

5 0
3 years ago
I need to know the right answer to that question
ad-work [718]
The answer is C 8.87*10^4 m/s (it shouldn't be m/s^2 though as velocity is in m/s)

Since you know the acceleration is 12 m/s^2, the initial velocity is 2.39*10^4 m/s and the time (you have to convert to seconds) is 5400 seconds, then you can use the equation

v = vo + at

When you plug in the values you get

v = 2.39*10^4 + 5400*12 . so v = 8.87*10^4 m/s. C is your answer.
8 0
4 years ago
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