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2 years ago

What is formula for distance

Physics

2 answers:

Georgia [21]2 years ago

7 0

Answer:

this is formula

Explanation:

Darina [25.2K]2 years ago

6 0

The formula for finding distance would be D = V x T ( distance = velocity multiplied by time )

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Does a heavier object or a lighter object experience a greater gravitational force?

aalyn [17]

L l Aigabeera ya q alr mma grasvdade l aferza uatare ms al amtaeialr lgerio hcaieno q edse pese masth

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2 years ago

Determine the kinetic energy of a 2000 kg roller coaster car that is moving at the speed of 10 ms

kondaur [170]

Answer:

$\boxed {\boxed {\sf 100,000 \ Joules}}$

Explanation:

Kinetic energy is energy due to motion. The formula is half the product of mass and velocity squared.

$E_k= \frac{1}{2} mv^2$

The mass of the roller coaster car is 2000 kilograms and the car is moving 10 meters per second.

m= 2000 kg
s= 10 m/s

Substitute these values into the formula.

$E_k= \frac{1}{2} (2000 \ kg ) \times (10 \ m/s)^2$

Solve the exponent.

(10 m/s)²= 10 m/s * 10 m/s= 100 m²/s²

$E_k= \frac{1}{2} (2000 \ kg ) \times (100 \ m^2/s^2)$

Multiply the first two numbers together.

$E_k= 1000 \ kg \times (100 \ m^2/s^2)$

Multiply again.

$E_k= 100,000 \ kg*m^2/s^2$

1 kilogram square meter per square second is equal to 1 Joule.
Our answer of 100,000 kg*m²/s² is equal to 100,000 Joules.

$E_k= 100,000 \ J$

The roller coaster car has 100,000 Joules of kinetic energy.

3 0

3 years ago

A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air

Bess [88]

Answer:

Explanation:

given,

initial velocity of the ball = 20 m/s

angle of ramp = 22°

ball travel at a distance = 5 m

a) for friction less

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh$

$v^2 = u^2 - 2gh$

$v = \sqrt{u^2- 2 g h cos 22^0}$

$v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0 }$

v = 17.58 m/s

b) considering the friction

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh-\mu_kmgl$

$v^2 = u^2 - 2gh-2\mu_kmgl$

$v = \sqrt{u^2- 2 g h cos 22^0-2\mu_kgl}$

$v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0-2\times 0.15\times 9.8 \times 5 }$

v = 17.16 m/s

7 0

3 years ago

jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh

Greeley [361]

Answer:

C. 590 mph

$\vert v_{cj}\vert=589.49\ mph$

Explanation:

Given:

velocity of jet, $v_j=500\ mph$
direction of velocity of jet, east relative to the ground
velocity of Cessna, $v_c=150\ mph$
direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

$\vec v_j=500\ \hat i\ mph$

$\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)$

$\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph$

Now the vector of relative velocity of Cessna with respect to jet:

$\vec v_{cj}=\vec v_j-\vec v_c$

$\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )$

$\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph$

Now the magnitude of this velocity:

$\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}$

$\vert v_{cj}\vert=589.49\ mph$ is the relative velocity of Cessna with respect to the jet.

8 0

3 years ago

PE=?, m=.6kg, g=10m/s2, h=35m PLS HELP I NEED THIS DONE

Karo-lina-s [1.5K]

210J

PE is mgh in this context.

7 0

4 years ago

What is formula for distance​

What is formula for distance