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Delicious77 [7]
3 years ago
6

A ball rolls down a hill, starting from rest. How long is it rolling if it accelerates at 3m/s2 and ends with a velocity of 35m/

s?
Physics
1 answer:
Kryger [21]3 years ago
6 0
<h2>Answer</h2>

The answer to this question is 11.67 s

<h2>Explanation</h2>

As we know that accelartion is the rate of change of velocity. So, it can be write as

a = (V_f -V_i) /t

where

V_f is the final velocity

V_i is the initial velocity

t is the time

a is the accelartion

as we konw

a = 3 ms^{-2}

t = ?

From rest So,

V_i = 0

V_f = 35 ms^{-1}

Putting values

3 = (35 - 0)/t

3 = 35/t

t = 35/3

t = 11.67 s

So, the right answe is 11.67 s

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Need help on this,<br>will mark brainliest if you answer.<br>​
Alecsey [184]

Answer:

(1) tropical storm

(2) Severe tropical storm

(3) 305 kmh

5 0
3 years ago
A cat jumps off a table which is 1.5 m high. If the initial velocity of the cat is 10 m/s, at an angle of 37 degrees above the h
Black_prince [1.1K]

Answer:

11.5 m

Explanation:

First, find the time it takes to land.

Given:

Δy = -1.5 m

v₀ = 10 sin 37° m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

-1.5 m = (10 sin 37° m/s) t + ½ (-9.8 m/s²) t²

-1.5 = 6.02 t − 4.9 t²

4.9 t² − 6.02 t − 1.5 = 0

Solve with quadratic formula:

t = [ -(-6.02) ± √((-6.02)² − 4(4.9)(-1.5)) ] / 2(4.9)

t = 1.44

Now find how far it moves horizontally in that time:

Given:

t = 1.44 s

v₀ = 10 cos 37° m/s

a = 0 m/s²

Find: Δx

Δx = v₀ t + ½ at²

Δx = (10 cos 37° m/s) (1.44 s) + ½ (0 m/s²) (1.44 s)²

Δx = 11.5 m

The cat lands 11.5 meters from the table.

7 0
3 years ago
If light is used to promote the electron, what are the selection rules for the initial and final spin and orbital angular moment
loris [4]

Answer:

(1)\Delta S=0

(2)\Delta L=0, +1,-1 but J=0 to J=0 transition not allowed.

Explanation:

Atoms can be described by the quantum number n, spin quantum number S, angular momentum quantum number L, and total angular momentum quantum number J. Based on approximation Russel- Saunders electron coupling, the atomic term symbol can be written as L_{J} ^{2S+1}.

The conditions or selection rule to promoting the electron are discussed below:

(1) The total spin should not change that is \Delta S=0.

(2) The total angular momentum change should be, \Delta L=0, +1,-1 but J=0 to J=0 transition not allowed.

8 0
3 years ago
If the absolute pressure inside the bottom of a container open to the atmosphere and filled with an unknown substance is 300 Pa.
anyanavicka [17]

Answer:

2.5 kg/m³

Explanation:

Absolute pressure = gauge pressure + atmospheric pressure

P = Pg + Pa

The gauge pressure caused by the weight of a fluid is called the static pressure.  It is equal to the density of the fluid × acceleration due to gravity × depth of the fluid,

Pg = ρgh

Therefore:

P = ρgh + Pa

300 Pa = ρ (10 m/s²) (8.00 m) + 100 Pa

ρ = 2.5 kg/m³

6 0
3 years ago
A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation. (a) Calculate the
professor190 [17]

Explanation:

It is given that,

A helicopter blade spins at exactly 100 revolutions per minute.

Its tip is 5.00 m from the center of rotation, r = 5 m

(a) Let v is the average speed of the blade tip in the helicopter’s frame of reference. Distance covered by the helicopter, d=2\pi r

In 100 revolutions, d=200\pi r=1000\pi

So, average speed of the blade tip in one second is given by :

v=\dfrac{d}{t}

v=\dfrac{1000\pi\ m}{60\ s}

v = 52.35 m/s

(b) The average velocity over one revolution is zero because the net displacement in one rotation is 0.

Hence, this is the required solution.

7 0
3 years ago
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