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vitfil [10]
3 years ago
5

A 0.2-kg steel ball is dropped straight down onto a hard, horizontal floor and bounces straight up. The ball's speed just before

and just after impact with the floor is 10 m/s. Determine the magnitude of the impulse delivered to the floor by the steel ball.
Physics
2 answers:
nadya68 [22]3 years ago
7 0

Answer:

Explanation:

Given

mass of steel ball m=0.2\ kg

initial speed of ball u=10\ m/s

Final speed of ball v=-10\ m/s (in upward direction)

Impulse imparted is given by change in the momentum of object

therefore impulse J is given by

J=\Delta P

\Delta P=m(v-u)

\Delta P=0.2(-10-10)

\Delta =-4\ N-s

so magnitude of Impulse =4 N-s

MAVERICK [17]3 years ago
4 0

Answer:

4 N s

Explanation:

mass, m = 0.2 kg

initial velocity, u = - 10 m/s (downward )

final velocity, v = + 10 m/s (upwards)

Impulse is defined a the change in momentum .

Impulse = m ( v - u)

Impulse = 0.2 ( 10 + 10)

Impulse = 4 N s

thus, the impulse is 4 N s .

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7 0
3 years ago
HELP PLS
melisa1 [442]

Answer:

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4 0
3 years ago
The musical tone a piano has a frequency of 346HZ and a wavelength of 1.3 M what is the speed of the sound
prohojiy [21]

Answer:

v= 449.8 m/s

Explanation:

Given data

Frequency=  346Hz

Wave length= 1.4m

The expression below is used to find the speed

v= f \lambda\\\\

substitute

v= 346*1.3\\\\v= 449.8 m/s

Hence the speed is v= 449.8 m/s

5 0
3 years ago
A heat pump with a COP of 3.15 is used to heat an air-tight house. When running, the heat pump consumes 5 kW of power. If the te
Jet001 [13]

Answer: 1026s, 17.1m

Explanation:

Given

COP of heat pump = 3.15

Mass of air, m = 1500kg

Initial temperature, T1 = 7°C

Final temperature, T2 = 22°C

Power of the heat pump, W = 5kW

The amount of heat needed to increase temperature in the house,

Q = mcΔT

Q = 1500 * 0.718 * (22 - 7)

Q = 1077 * 15

Q = 16155

Rate at which heat is supplied to the house is

Q' = COP * W

Q' = 3.15 * 5

Q' = 15.75

Time required to raise the temperature is

Δt = Q/Q'

Δt = 16155 / 15.75

Δt = 1025.7 s

Δt ~ 1026 s

Δt ~ 17.1 min

5 0
3 years ago
Read 2 more answers
A flutist assembles her flute in a room where the speed of sound is 342 m/s. When she plays the note A, it is in perfect tune wi
sertanlavr [38]

Answer:

5.15348 Beats/s

4.55 mm

Explanation:

v_1 = Velocity of sound = 342 m/s

v_2 = Velocity of sound = 346 m/s

f_1 = First frequency = 440 Hz

Frequency is given by

f_2=\frac{v_2}{2L_1}\\\Rightarrow f_2=\frac{346}{2\times 0.38863}\\\Rightarrow f_2=445.15348\ Hz

Beat frequency is given by

|f_1-f_2|=|440-445.15348|=5.15348\ Beats/s

Beat frequency is 5.15348 Hz

Wavelength is given by

\lambda_1=\frac{v_1}{f}\\\Rightarrow \lambda_1=\frac{342}{440}\\\Rightarrow \lambda_1=0.77727\ m

Relation between length of the flute and wavelength is

\lambda_1=2L_1\\\Rightarrow L_1=\frac{\lambda_1}{2}\\\Rightarrow L_1=\frac{0.77727}{2}\\\Rightarrow L_1=0.38863\ m

At v = 346 m/s

\lambda_2=\frac{v_2}{f}\\\Rightarrow \lambda_2=\frac{346}{440}\\\Rightarrow \lambda_1=0.78636\ m

L_2=\frac{\lambda_2}{2}\\\Rightarrow L_2=\frac{0.78636}{2}\\\Rightarrow L_2=0.39318\ m

Difference in length is

\Delta L=L_2-L_1\\\Rightarrow \Delta L=0.39318-0.38863\\\Rightarrow \Delta L=0.00455\ m=4.55\ mm

It extends to 4.55 mm

7 0
3 years ago
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