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ruslelena [56]
3 years ago
12

LOADS OF POINTS - PLEASE SHOW WORKING AND METHOD USED

Physics
2 answers:
erastova [34]3 years ago
8 0
The formula your looking for is Vf = Vi + A (T)

Vf = Velocity Final
Vi = Velocity Initial
A = Acceleration
T = Time

Our Initial Velocity is 15m/s. That’s our starting point.

Acceleration is -3m/s. It’s negative because we’re slowing down

We know that the cyclist comes to a complete stop, which means the Final Velocity is 0

In order to solve this with the equation. We need three of the four variables. We just don’t know time. So our setup is -

0 = 15-3(T)

Then you solve

NeTakaya3 years ago
4 0

Answer:hi

Explanation:

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suppose you want to determine the surface area of this sugar cube. it has edges that are each 2 cm long. if you cut the cube in
ivolga24 [154]

Answer:

Half: 6 cm^2    Whole: 12 cm^2

Explanation:

First, we know that the edges of the cube are 2 cm long. So there are 6 faces on a cube. We do 2x6=12 cm^2 as our total surface area. Then it asks for each half. So you would divide it by 2 and get 6 cm^2 as your half.

5 0
3 years ago
A roller coaster travels down a 120 m track in 12.5 seconds how fast does the roller coaster go
KIM [24]

Answer:

9.6m/s

Explanation:

Using the equation S=d/t where s=speed, d=distance, and t=time

plug in the known variables

S=120m/12.5s

S=9.6m/s

4 0
3 years ago
(a) Calculate the magnitude of the gravitational force exerted by Mars on a 80 kg human standing on the surface of Mars. (The ma
andrew11 [14]

Answer:

a) F=1.044\times 10^9\ N

b)F'=1.044\times 10^9\ N

c) F_p=1.0672\times10^{-7}\ N

d) Treat the humans as though they were points or uniform-density spheres.

Explanation:

Given:

  • mass of Mars, M=6.4\times 10^{23}\ kg
  • radius of the Mars, r=3.4\times 10^{6}\ m
  • mass of human, m=80\ kg

a)

Gravitation force exerted by the Mars on the human body:

F=G.\frac{M.m}{r^2}

where:

G=6.67 \times 10^{-11}\ m^3.kg^{-1}.s^{-2} = gravitational constant

F=6.67\times10^{-11}\times \frac{6.4\times 10^{23}\times 80}{(3.4\times 10^{6})^2}

F=1.044\times 10^9\ N

b)

The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.

F'=F

F'=1.044\times 10^9\ N

c)

When a similar person of the same mass is standing at a distance of 4 meters:

F_p=6.67\times10^{-11}\times \frac{80\times 80}{4}

F_p=1.0672\times10^{-7}\ N

d)

The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.

  • Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.
  • Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.
4 0
2 years ago
A cannonball is fired perfectly horizontally from the top of a 210 m tall cliff. It is fired with an initial velocity of 50 m/s.
pochemuha

Answer:

the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

Explanation:

Given;

height of the cliff, h = 210 m

initial horizontal velocity of the cannonball, Ux = 50 m/s

initial vertical velocity of the cannonball, Uy = 0

The time for the cannonball to reach the ground is calculated as;

h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t  = 6.55 \ s

The horizontal distance covered by the cannonball before it hits the ground is calculated as;

X = U_x \times \ t\\\\X = 50 \times \ 6.55\\\\X = 327.5 \ m

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

8 0
2 years ago
What mediums are best for light waves?
ycow [4]
The strength of the electric and magnetic fields there is no physical "distance" of oscillation here. nothing is actually moving up and down if you draw light as a sinusoidal wave, the up and down motion is the strength of the EM fields cheers
4 0
3 years ago
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