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Bezzdna [24]
3 years ago
12

Mass of 55 kg riding in a car at20m/s slams on brakes rests in.5 sec how much force from seatbelt

Physics
1 answer:
Evgen [1.6K]3 years ago
5 0
2,200 Newtons.

F = ma

v f = v i + at
0mls = 20mls + a(.5s)

-20mls/.5s = -40mls squared

-40mls squared (55kg) = 2,200 Newtons.

Hope this helps!
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Which type of reaction occurs when you bake a tray of cookies?
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Answer:

b

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6 0
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A car accelerates uniformly from rest to a speed of 7.6 m/s in 4.6 s. How far does the car travel in that time?
uysha [10]

Answer:

The car will travel a distance of 17.45 meters.

Explanation:

Given:

Initial velocity (V_i) = 0

Final velocity (V_f) = 7.6 m/s

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Acceleration = (Final velocity - Initial Velocity )/time

a=\frac{(V_f-Vi)}{t}= \frac{7.6-0}{4.6}=1.65\ ms^-2

We have to calculate total distance traveled by the car.

Let the distance traveled be 'd'

Equation of motion:

d=V_i(t)+\frac{at^2}{2}

Plugging the values.

⇒d=V_i(t)+\frac{at^2}{2}

⇒d=0+\frac{1.65*(4.6)^2}{2}

⇒d=17.45\ m

The car will travel a distance of 17.45 meters for the above case.

4 0
3 years ago
an 269 kg object is moved a distance of 1.9 m by a force if 580 j of work is done on the object what is the object acceleration
IrinaK [193]

Answer:

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7 0
3 years ago
An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens. a) What is the focal length
PSYCHO15rus [73]

Answer:

-2.25

Real and inverted image

Explanation:

u = Object distance =  16 cm

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Lens Equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{16}+\frac{1}{36}\\\Rightarrow \frac{1}{f}=\frac{13}{144}\\\Rightarrow f=\frac{144}{13}=11.07\ cm

Focal length of the lens is 11.07 cm. Positive value indicates the lens is a convex lens

Magnification

m=-\frac{v}{u}\\\Rightarrow m=-\frac{36}{16}\\\Rightarrow m=-2.25

Since magnification is negative the image inverted

3 0
3 years ago
How do I do trebuchet calculations????? Help me please
Mnenie [13.5K]
I found some good web pages with highly detailed answers to predicting the range of a trebuchet. A very simple model we have used in my Intro to Eng class just uses the mass of the projectile (m2), the mass of the counter weight (m1), and the height the counter weight falls (h): Range (max) = 2 * (m1/m2) * h Now the efficiency of the trebuchet will cause this model to be off by quite a bit. But once you have a working trebuchet, we find this model works well when we vary m1, m2, or h. We assume we have a take off angle of 45 degrees above the horizon. This solution is based on the classic max range ballistics problem - 45 degree take off angle. It also assumes converting all the potential energy of the counter weight to kinetic energy of the projectile. That is why the efficiency issue comes up as a lot of energy is lost due to friction in the moving trebuchet. If the projectile spins a lot then it will travel a shorter distance as the potential energy is split into kinetic and rotational energy. Projectile shape and wind will also vary the results. Good luck.
5 0
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