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3 years ago

12

A stone is thrown downward with a speed of 8 m/s from a height of 14 m. (acceleration due to gravity: 10 m/s2) What is the speed

(in m/s) of the stone just before it hits the ground

1 answer:

max2010maxim [7]3 years ago

6 0

Answer:

The speed of the stone just before it hits the ground is 18.54 m/s

Explanation:

Given that,

Initial speed of the stone, u = 8 m/s

The stone is thrown downward from a height of 14 m

We need to find the speed of the stone just before it hits the ground. It can be calculated using third equation of motion as :

$v^2-u^2=2ah$

v is the speed of the stone just before it hits the ground

$v^2=2ah+u^2$

$v^2=2\times 10\times 14+(8)^2$

v = 18.54 m/s

So, the speed of the stone just before it hits the ground is 18.54 m/s. Hence, this is the required solution.

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A cross country skier can complete a 7.5km race in 45 minutes. What is the skier's average speed?

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0.3 hope this is right.

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3 years ago

A compressed gas cylinder containing 1.50 mol methane has a volume of 3.30 L. What pressure does the methane exert on the walls

sweet-ann [11.9K]

Answer:

P=11.126 atm

Explanation:

We are given that a compressed gas cylinder containing 1.50 mol methane

Volume of methane in a gas cylinder =3.3 L

Temperature = $25^{\circ}=25+273.15=298.15 K$

We have to find the pressure exert by gas on cylinder

Using ideal gas equation

PV =nRT

Where R= gas constant=0.0821 L-atm/K mol

Substitute all given values

Then we get

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$P=\frac{36.717}{3.3}$

P=11.126 atm

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4 years ago

Read 2 more answers

A shuttle bus slows down with an average acceleration of -2.4 m/s2. How long does it

olganol [36]

Answer:

$\boxed {\boxed {\sf 3.75 \ seconds }}$

Explanation:

Average acceleration is found by dividing the change in acceleration by the time.

$a=\frac{ v_f-v_i}{t}$

The shuttle bus has an acceleration of -2.4 meters per square second. It slows from 9.0 meters per second to rest, or 0 meters per second. Therefore:

$a= -2.4 \ m/s^2 \\v_f= 0 \ m/s \\v_i= 9 \ m/s$

Substitute the values into the formula.

$-2.4 \ m/s^2=\frac{0 \ m/s - 9 \ m/s}{t }$

Solve the numerator.

$-2.4 \ m/s^2 = \frac{-9 \ m/s}{t}$

We want to solve for t, the time. We have to isolate the variable. Let's cross multiply.

$\frac{-2.4 \ m/s^2}{1} = \frac{-9 \ m/s}{t}$

$-9 \ m/s *1= -2.4 \ m/s^2 *t$

$-9 \ m/s=-2.4 m/s^2*t$

t is being multiplied by -2.4. The inverse of multiplication is division, so divide both sides by -2.4

$\frac{-9 \ m/s }{-2.4 \ m/s^2} =\frac{ -2.4 \ m/s^2*t}{-2.4 \ m/s^2}$

$\frac{-9 \ m/s }{-2.4 \ m/s^2} =t$

$3.75 \ s=t$

It takes <u>3.75 seconds.</u>

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3 years ago

A cylindrical rod 21.5 cm long with a mass of 1.20 kg and a radius of 1.50 cm has a ball of diameter of 6.90 cm and a mass of 2.

Sergeeva-Olga [200]

Answer

given,

length of rod = 21.5 cm = 0.215 m

mass of rod (m) = 1.2 Kg

radius, r = 1.50

mass of ball, M = 2 Kg

radius of ball, r = 6.90/2 = 3.45 cm = 0.0345 m

considering the rod is thin

$I = \dfrac{1}{3}M_{rod}L^2 + [\dfrac{2}{5}M_{ball}R^2+M_{ball}(R+L)^2]$

$I = \dfrac{1}{3}\times 1.2 \times 0.215^2 + [\dfrac{2}{5}\times 2 \times 0.0345^2+2\times (0.0345 +0.215)^2]$

I = 0.144 kg.m²

rotational kinetic energy of the rod is equal to

$KE = M_{rod}g\dfrac{L}{2} + M_{ball}g(L+R)^2$

$KE = 1.2 \times 9.8 \times \dfrac{0.215}{2} + 2\times 9.8\times (0.215+0.0345)^2$

KE = 6.15 J

b) using conservation of energy

$K_f + U_f = K_i + U_i + \Delta E$

$\dfrac{1}{2}I\omega^2+ 0=0 + 6.15+0$

$\dfrac{1}{2}\times 0.144 \times \omega^2= 6.15$

ω = 9.25 rad/s

c) linear speed of the ball

v = r ω

v = (L+R )ω

v = (0.215+0.0345) x 9.25

v =2.31 m/s

d) using equation of motion

v² = u² + 2 g h

v² = 0 + 2 x 9.8 x 0.248

v = √4.86

v =2.20 m/s

speed attained by the swing is more than free fall

% greater = $\dfrac{2.31-2.20}{2.20}\times 100$

= 5 %

speed of swing is 5 % more than free fall

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3 years ago

A 47-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37 degrees above the

levacccp [35]

Tension in the rope due to applied force will be given as

$F = 142 N$

angle of applied force with horizontal is 37 degree

displacement along the floor = 6.1 m

so here we can use the formula of work done

$W = F d cos\theta$

now we can plug in all values above

$W = 142 * 6.1 * cos37$

$W = 691.8 J$

So here work done to pull is given by 691.8 J

8 0

3 years ago