A roller coaster goes from 10 m/s to 30 m/s in 5 seconds. What is it's acceleration? <br> Vf-Vi/T

A ball is thrown at 20 m/s from the ground upwards at an angle of elevation of 30°. How far away does it land? 35.35 m

mestny [16]

Answer:

35.35 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 20 m/s

Angle of projection (θ) = 30°

Acceleration due to gravity (g) = 9.8 m/s²

Range (R) =.?

The range (i.e how far away) of the ball can be obtained as follow:

R = u² Sine 2θ /g

R = 20² Sine (2×30) / 9.8

R = 400 Sine 60 / 9.8

R = (400 × 0866) / 9.8

R = 346.4 / 9.8

R = 35.35 m

Therefore, the range (i.e how far away) of the ball is 35.35 m

8 0

2 years ago

If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite d

Yuliya22 [10]

Incomplete question as the mass of baseball is missing.I have assume 0.2kg mass of baseball.So complete question is:

A baseball has mass 0.2 kg.If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Answer:

ΔP=20 kg.m/s

Explanation:

Given data

Mass m=0.2 kg

Initial speed Vi=-44.5m/s

Final speed Vf=55.5 m/s

Required

Change in momentum ΔP

Solution

First we take the batted balls velocity as the final velocity and its direction is the positive direction and we take the pitched balls velocity as the initial velocity and so its direction will be negative direction.So we have:

$v_{i}=-44.5m/s\\v_{f}=55.5m/s$

Now we need to find the initial momentum

So

$P_{1}=m*v_{i}$

Substitute the given values

$P_{1}=(0.2kg)(-44.5m/s)\\P_{1}=-8.9kg.m/s$

Now for final momentum

$P_{2}=mv_{f}\\P_{2}=(0.2kg)(55.5m/s)\\P_{2}=11.1kg.m/s$

So the change in momentum is given as:

ΔP=P₂-P₁

$=[(11.1kg.m/s)-(-8.9kg.m/s)]\\=20kg.m/s$

ΔP=20 kg.m/s

3 0

2 years ago

An object is thrown upward from the top of a 144-foot building with an initial velocity of 128 feet per second. The height h of

ankoles [38]

Answer:

After 9 seconds the object reaches ground.

Step-by-step explanation:

We equation of motion given as h = -16t²+128t+144,

We need to find in how many seconds will the object hit the ground,

That is we need to find time when h = 0

0 = -16t²+128t+144

16t²-128t-144= 0

$t=\frac{-(-128)\pm \sqrt{(-128)^2-4\times 16\times (-144)}}{2\times 16}\\\\t=\frac{128\pm \sqrt{25600}}{32}\\\\t=\frac{128\pm 160}{32}\\\\t=9s\texttt{ or }t=-1s$

Negative time is not possible, hence after 9 seconds the object reaches ground.

8 0

2 years ago

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

muminat

Answer:

a) E = 0

b) $E = \dfrac{k_e \cdot q}{ r^2 }$

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

From which we have;

$E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0$

E = 0/A = 0

E = 0

b) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

$E \cdot A = \dfrac{+q }{\varepsilon _{0}}$

$E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}$

By Gauss theorem, we have;

$E\oint dS = \dfrac{q}{\varepsilon _{0}}$

Therefore, we get;

$E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}$

The electrical field outside the spherical shell

$E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }$

$k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }$

Therefore, we have;

$E = \dfrac{k_e \cdot q}{ r^2 }$

5 0

2 years ago

PLEASE HELP ASAP 36. The Doppler Effect applies to light in addition to sound. What do you call the phenomenon that causes dista

skad [1K]

Answer:

It's probably the Redshift and Blueshift

Explanation:

The light shifting towards shorter or longer wavelengths as objects in space (stars or galaxies) move closer or farther away from us. This phenomenon is known as the Redshifts and Blueshifts. The concept indicates us that the Universe is expanding.

Hope this helps you!

Bye!

3 0

2 years ago

A roller coaster goes from 10 m/s to 30 m/s in 5 seconds. What is it's acceleration? Vf-Vi/T