What does each letter stand for in the formula p=v x i?

5. No matter what month and date your birthday is on, NASA says it has a significant picture from

padilas [110]

Answer:

The Hubble space telescope.

Explanation:

Hubble is a telescope that observers the sky 24/7 non-stop, which means that for every day of the year it would have made a significant discovery, which of course includes your birthday. Furthermore, you can actually go to NASA website and find out what discovery was made on your birthday! This shows both the vastness of the universe (it really has to be huge for a telescope to have a discovery for each day of the year!) and the ceaseless work of the telescope!

7 0

3 years ago

Find an expression for the kinetic energy of the car at the top of the loop.Express the kinetic energy in terms of m, g, h, and

lyudmila [28]

Answer:

K.E₂ = mg(h - 2R)

Explanation:

The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:

K.E₁ + P.E₁ = K.E₂ + P.E₂

where,

K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)

P.E₁ = Initial Potential Energy = mgh

K.E₂ = Final Kinetic Energy at the top of the loop = ?

P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)

Therefore,

0 + mgh = K.E₂ + mg(2R)

K.E₂ = mg(h - 2R)

3 0

3 years ago

An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1

goblinko [34]

Answer:

$v"_{1} = v_{1} tan$ Θ

$v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

Θ = $tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )$

Explanation:

Applying the law of conservation of momentum, we have:

Δ $p_{x = 0}$

$p_{x} = p"_{x}$

$m_{1}v_{1} = m_{2}v"_{2} cos$ Θ (Equation 1)

Δ $p_{y} = 0$

$p_{y} = p"_{y}$

$0 = m_{1} v"_{1} - m_{2} v"_{2} sin$ Θ (Equation 2)

From Equation 1:

$v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

From Equation 2:

$m_{2} v"_{2}$ sinΘ = $m_{1} v_{1}$

$v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }$

Replacing Equation 3 in Equation 4:

$v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}$

$v"_{1}=v_{1}\frac{sinΘ}{cosΘ}$

$v"_{1}=v_{1}tan$ Θ (Equation 5)

And we found Θ from the Equation 5:

tanΘ= $\frac{v"_{1}}{v_{1}}$

Θ= $tan^{-1}(\frac{v"_{1}}{v_{1}})$

7 0

3 years ago

A rugby player sits on a scrum machine that weighs 200 Newtons. Given that the coefficient of static friction is 0.67, the coeff

Trava [24]

a. 850 N is the minimum force needed to get the machine/player system moving, which means this is the maximum magnitude of static friction between the system and the surface they stand on.

By Newton's second law, at the moment right before the system starts to move,

• net horizontal force

∑ F[h] = F[push] - F[s. friction] = 0

• net vertical force

∑ F[v] = F[normal] - F[weight] = 0

and we have

F[s. friction] = µ[s] F[normal]

It follows that

F[weight] = F[normal] = (850 N) / (0.67) = 1268.66 N

where F[weight] is the combined weight of the player and machine. We're given the machine's weight is 200 N, so the player weighs 1068.66 N and hence has a mass of

(1068.66 N) / g ≈ 110 kg

b. To keep the system moving at a constant speed, the second-law equations from part (a) change only slightly to

∑ F[h] = F[push] - F[k. friction] = 0

∑ F[v] = F[normal] - F[weight] = 0

so that

F[k. friction] = µ[k] F[normal] = 0.56 (1268.66 N) = 710.45 N

and so the minimum force needed to keep the system moving is

F[push] = 710.45 N ≈ 710 N

4 0

1 year ago

A curve that has a radius of 90 m is banked at an angle of =10.8∘. If a 1100 kg car navigates the curve at 75 km/h without skidd

PilotLPTM [1.2K]

The minimum coefficient of static friction between the pavement and the tires is 0.69.

The given parameters;

radius of the curve, r = 90 m
angle of inclination, θ = 10.8⁰
speed of the car, v = 75 km/h = 20.83 m/s
mass of the car, m = 1100 kg

The normal force on the car is calculated as follows;

$F_n = mgcos(\theta)$

The frictional force between the car and the road is calculated as;

$F_k = \mu_k F_n\\\\F_k = \mu_k mgcos(\theta)$

The net force on the car is calculated as follows;

$mgsin(\theta) + \mu_s mgcos(\theta) = \frac{mv^2}{r} \\\\mg(sin\theta \ + \ \mu_s cos\theta)= \frac{mv^2}{r} \\\\g(sin\theta \ + \ \mu_s cos\theta)= \frac{v^2}{r}\\\\sin\theta \ + \ \mu_s cos\theta = \frac{v^2}{rg}\\\\\mu_s cos\theta = sin\theta \ + \ \frac{v^2}{rg}\\\\\mu_s = \frac{sin\theta}{cos \theta} + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(\theta) + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(10.8) + \frac{(20.83)^2}{cos(10.8) \times 90 \times 9.8} \\\\\mu_s = 0.19 + 0.5\\\\$

$\mu_s = 0.69$

Thus, the minimum coefficient of static friction between the pavement and the tires is 0.69.

Learn more here:brainly.com/question/15415163

8 0

2 years ago