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ElenaW [278]
3 years ago
15

What are the 3 formulas you can use for vertical motion for a projectile?

Physics
1 answer:
klio [65]3 years ago
4 0
I hope this can help you ask me if you need help again

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Which statement is true for a sound wave entering an area of warmer air
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That waves travel faster than the wave lenght!
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3 years ago
Read 2 more answers
the velocity of an object is given v(t), and the acceleration is given by a(t). what is the relationship between the total
zmey [24]

The rate at which velocity changes is called acceleration. (Attensity exists when velocity varies.) If a moving object changes speed.

Why does time accelerate the rate at which velocity changes?

A motion's acceleration is the rate at which it changes from one velocity to another. A velocity's rate of change with respect to time is referred to as its acceleration. The amount and direction of acceleration are both properties of a vector quantity.

A change in velocity is known as what?

A velocity change's acceleration is measured. Acceleration is the measure of how quickly a velocity changes with time. The acceleration measure used in SI is M/s2.

To know more about velocity visit: brainly.com/question/18084516?

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8 0
1 year ago
A train that has wheels with a diameter of 91.44 cm (36 inches used for 100 ton capacity cars) slows down from 82.5 km/h to 32.5
podryga [215]

Answer:

The answer is below

Explanation:

The initial velocity = u = 82.5 km/h = 22.92 m/s, the final velocity = 32.5 km/h = 9.03 m/s, diameter = 91.55 cm = 0.9144 cm

radius (r) = diameter / 2 = 0.9144 / 2= 0.4572 m

a) Initial angular velocity (\omega_o) = u /r = 22.92 / 0.4572 = 50.13 rad/s, final velocity  (ω) = v / r = 9.03 / 0.4592 = 19.67 rad / s

θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

angular acceleration (α) is:

\omega^2=\omega_o^2+2\alpha \theta\\\\19.67^2-50.13^2=2\alpha(272.9)\\\\19.67^2=50.13^2+2\alpha(272.9)\\\\2\alpha(272.9)=-2126.108\\\\\alpha=-3.89\ rad/s^2\\\\

b)

\omega=\omega_o+\alpha t\\\\19.67=50.13+(-3.89t)\\\\3.89t=50.13-19..67\\\\3.89t=30.46\\\\t=7.83\ s

c) θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

a) When it stops, the final angular velocity is 0. Hence:

\omega^2=\omega_o^2+2\alpha \theta\\\\0=50.13^2+2(-3.89)\theta\\\\2(3.89)\theta=50.13^2\\\\2(3.89)\theta=2513\\\\\theta=323\ rad\\\\revolutions=\frac{\theta}{2\pi r}=\frac{323}{2\pi(0.4572)}  =112.4\ rev

θ = 323 rad

4 0
3 years ago
N
PtichkaEL [24]
Hi I need help with some questions I have on a quiz / test
4 0
3 years ago
A block of mass 57.1 kg rests on a slope having an angle of elevation of 28.3°. If pushing downhill on the block with a force ju
stira [4]

Answer:

The coefficient is 0.90

Explanation:

Drawing a diagram makes thing easier, we will assume that the acceleration tends to zero because it start barely moving.

-F_s+mg*sin(\theta)+F=0\\F_s=57.1kg*9.8m/s^2*sin(28.3)+177N\\F_s=442N\\F_s=\µ*N\\N=m*g*cos(\theta)\\N=57.1*9.8*cos(28.3)=493N\\\\\µ=\frac{442N}{493N}=0.90

3 0
3 years ago
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