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valentina_108 [34]
2 years ago
8

A 1.0 kg ball falls from rest a distance of 19.6 m.

Physics
1 answer:
Nimfa-mama [501]2 years ago
6 0

Answer:

192.08J

19.6m/s

Explanation:

Since there will be no potential energy when the ball is on the ground, the change in potential energy is equal to the potential energy at the start when the ball is 19.6m above the ground.

PE=mgh

=(1)(9.8)(19.6)

=192.08J

v²=u²+2as, where v is the final velocity, u is initial velocity, a is acceleration and s is distance. Initial velocity is 0 since it starts at rest.

v²=u²+2as

v²=0²+2(9.8)(19.6)

v=√384.16

=19.6m/s

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One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor
Bess [88]

Answer:

a) The final velocity is 20 m/s when the large-mass object is the one moving initially.

b) The final velocity is 9.0 m/s when the small-mass object is the one moving initially.

Explanation:

The momentum of the system is calculated as the sum of the momenta of each object. Each momentum is calculated as follows:

p = m · v

Where:

p =  momentum.

m =  mass.

v = velocity.

Then, the momentum of the system is the following:

m1 · v1 + m2 · v2 = (m1 + m2) · v

Where:

m1 = mass of the bigger object.

v1 = velocity of the bigger object.

m2 = mass of the smaller object.

v2 = velocity of the smaller object.

v = final velocity of the two objects after the collision.

Solving the equation for the final velocity:

(m1 · v1 + m2 · v2)/ (m1 + m2) = v

a) Let´s calculate the final velocity when the bigger object is moving:

(7.1 kg · 29 m/s + 3.2 kg · 0)/(7.1 kg + 3.2 kg) = v

<u>v = 20 m/s</u>

b) When the smaller object is moving:

(7.1 kg · 0 m/s + 3.2 kg · 29 m/s) / (7.1 kg + 3.2 kg) = v

<u>v = 9.0 m/s</u>

7 0
3 years ago
Find the wavelength λ of the 80.0-khz wave emitted by the bat. express your answer in millimeters.
vfiekz [6]

Answer:

4.29 millimeters

Explanation:

Bats emit ultrasound waves: in air, ultrasound waves travel at a speed of

v=343 m/s

The frequency of the waves emitted by this bat is:

f=80.0 kHz = 80,000 Hz

Therefore we can find the wavelength of the wave emitted by the bat by using the relationship between speed, frequency and wavelength:

\lambda=\frac{v}{f}=\frac{343 m/s}{80,000 Hz}=4.29\cdot 10^{-3} m=4.29 mm

4 0
3 years ago
In which situation is work not being done?
almond37 [142]

AS

work done =W = F.d = F d cosФ     (Ф is angle between force F and displacement d) If a body/object is moving on a smooth surface (friction-less surface ) .There is no force acting on that body.  F=0 so W=FdcosФ= (0)dcosФ ⇒ W=0

Now if a body is facing some amount of force but under the action of force there is no displacement covered. d=0 so W =FdcosФ= F(0)cosФ ⇒W=0

example:  A person is applying a force on rigid wall but wall remains at rest there is no displacement occurs in wall.

The third term upon which work done  dependent is angle between force and displacement i.e Ф. If Ф=90° then W= FdcosФ= Fdcos90⇒ W=0   ( as cos 90°=0)

5 0
3 years ago
The boiling point of water is 91.30 °C on the
Taya2010 [7]

Answer:

196.34 °F

Explanation:

To convert from degrees celsius to degrees fahrenheit, use this equation:

(°C * 9/5) + 32 = °F

So, using this equation:

(91.30 * 9/5) + 32 = °F

196.34 + 32 = °F

°F = 196.34

Hope this helps!

3 0
3 years ago
The volume flow rate of the water supplied by a well is 2.0×10−4m3/s.The well is 40.0 m deep. (a) What is the power output of th
MaRussiya [10]

Answer:

a). P=78.4W

b). P=392kPa

c.) It must be at the bottom

Explanation:

Given:

Volume flow V_f=2.0x10^{-4}m^3/s

Well depp h=40.m

a.

The power output of the pum

W=F*d

F=m*g

m=p*V=1000kg/m^3*2.0x10^{-4}m^3}=0.2Kg

W=m*g*d=0.2kg*9.8m/s^2*40.0m=78.4kg*m^2/s^2

W=78.4J

P=\frac{W}{t}=\frac{78.4J}{1s}=78.4W

b.

The pressure of difference the pum

ΔP=p*g*y'

ΔP=1000kg/m^3*9.8m/s^2*40.0m=392x10^3Pa

P=392kPa

c.

It must be at the bottom since the pressure difference is greater than atmospheric pressure, so it wouldn't be able to lift the water all the way  

4 0
3 years ago
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