What is the cheetahs momentum?

Cars driving 15.7 m/s on a flat ground drives off of 25.3 m High cliff how far from the base of the cliff does it land

MA_775_DIABLO [31]

The car lands 35.6 m from the base of the cliff

Explanation:

The motion of the car is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

We start by analyzing the vertical motion, in order to find the time of flight of the car. We do it by using the following suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where:

s = 25.3 m is the vertical displacement of the car (the height of the cliff)

$u_y=0$ is the initial vertical velocity of the car (because it is moving horizontally)

t is the time of the fall

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find :

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(25.3)}{9.8}}=2.27 s$

Now we can analyze the horizontal motion of the car, which moves with constant horizontal velocity of

$v_x = 15.7 m/s$

Therefore, the horizontal distance covered in a time t is

$d=v_x t$

and by substituting t = 2.27 s, we find the how far from the base of the cliff the car lands:

$d=(15.7)(2.27)=35.6 m$

Learn more about projectile motion:

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7 0

3 years ago

Change the value of the battery emf to 10.0 V, and make sure the middle resistor is set to 20 Ω. Use extra wire for each "leg" o

Inga [223]

Answer:

The current increases when the circuit is closed.

Explanation:

As the complete question is not given, the complete question is attached herewith.

From the data the resistors are connected in parallel and the value of Battery EMF is increased thus as the resistance is decreased and the EMF of the battery is increased, thus the current will increase when the switch is closed.

6 0

4 years ago

Read 2 more answers

400 gallons of pesticide is accidentally spilled into a lake and uniformly mixes with the water. The volume of the lake includin

svet-max [94.6K]

Answer:

t=27725.8minutes

Explanation:

To determinate use the law the change of volume in determinate time is derivate so:

$\frac{dV}{dt}=flowin-flowout$

The flow in is zero and the flow out is modeling with V as the volume out in the volume of the lake in the reason of change.

$\frac{dV}{dt}=0-(\frac{V}{10x10^{8}})*5000\\\frac{dV}{dt}=-5x10^{-5}*V \\dV*V=-5x10^{-5}*dt$

Integrate to get the volume function

$\int\limits{\frac{1}{V}}\, dv=\int\limits{-5x10^{-5} }\, dt\\ ln(V)=-5x10^{-5}*t\\ln*e(V)=C*e^{-5x10^{-5}*t}\\$

The find the constant C in the time t=0 the volume is knowing so:

$t=0s\\V=400\\V=C*e^{0}\\C=400$

$V=400*e^{-5x10^{-5}*t}$

The volume is 1 part per million so:

$\frac{V}{10x^{8}}=\frac{1}{10x^{6}}\\ V=100$

$V=400*e^{-5x10^{-5}*t}\\100=400*e^{-5x10^{-5}*t}\\\frac{100}{400}=e^{-5x10^{-5}*t}\\ln*\frac{1}{4}=ln*e^{-5x10^{-5}*t}\\-ln(4)=-5x10^{-5}*t\\ t=\frac{-ln(4)}{-5x10^{-5}}=27725.88\\t=27725.8minutes$

8 0

3 years ago

How strong an electric field is needed to accelerate electrons in an X-ray tube from rest to one-tenth the speed of light in a d

Bond [772]

Answer:

E= 50.1*10³ N/C

Explanation:

Assuming no other forces acting on the electron, if the acceleration is constant, we can use the following kinematic equation in order to find the magnitude of the acceleration:

$vf^{2} -vo^{2} = 2*a*x$

We know that v₀ = 0 (it starts from rest), that vf = 0.1*c, and that x = 0.051 m, so we can solve for a, as follows:

$a = \frac{vf^{2}}{2*x} = \frac{(3e7 m/s)^{2} }{2*0.051m} =8.8e15 m/s2$

According to Newton's 2nd Law, this acceleration must be produced by a net force, acting on the electron.

Assuming no other forces present, this force must be due to the electric field, and by definition of electric field, is as follows:

F = q*E (1)

In this case, q=e= 1.6*10⁻19 C

But this force, can be expressed in this way, according Newton's 2nd Law:

F = m*a (2) ,

where m= me = 9.1*10⁻³¹ kg, and a = 8.8*10¹⁵ m/s², as we have just found out.

From (1) and (2), we can solve for E, as follows:

$E=\frac{me*a}{e} =\frac{(9.1e-31 kg)*(8.8e15m/s2)}{1.6e-19C} = 50.1e3 N/C$

⇒ E = 50.1*10³ N/C

3 0

4 years ago

A car traveling 20 m/s to the north collides with a car traveling 10 m/s to the

sergiy2304 [10]

Answer: C

Explanation:

In collision, whether elastic or inelastic collisions, momentum is always conserved. That is, the momentum before collision will be equal to the momentum after collision.

Change in momentum of the system will be momentum after collision minus total momentum before collision.

Since momentum is a vector quantity, the direction will also be considered.

Momentum = MV - mU

Let

M = 800 kg is going north

at V = 20 m/s and the other car

m= 800 kg is going south

at U = 10m/s.

Substitute all the parameters into the formula

Momentum = (800 × 20) - (800 × 10)

= 8000 kgm/s

The final momentum after collision will also be equal to 8000 kgm/s

Change in momentum = 8000 - 8000

Change in momentum = 0

5 0

3 years ago

Read 2 more answers