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mafiozo [28]
3 years ago
10

What is the momentum of a 0.1-kg mass moving with a speed of 5 m/s

Physics
1 answer:
Andrej [43]3 years ago
4 0
This question is wrong because in momontum we will write acceleration instead of speed.     suppose acceleration is 5m/s2 then 
 P= ma 
                then put values
   
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If a car is moving backward and has negative acceleration, what can be said about the speed of the car?
KATRIN_1 [288]
I think its A let me know its wrong or not
4 0
3 years ago
Read 2 more answers
A ball is launched with initial speed v from ground level up a frictionless slope (This means the ball slides up the slope witho
amid [387]

Answer:

hmax = 1/2 · v²/g

Explanation:

Hi there!

Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.

KE = PE

Where KE is the initial kinetic energy and PE is the final potential energy.

The kinetic energy of the ball is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass of the ball

v = velocity.

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the ball.

g = acceleration due to gravity (known value: 9.81 m/s²).

h = height.

At  the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:

PE = KE

m · g · hmax = 1/2 · m · v²

Solving  for hmax:

hmax = 1/2 · v² / g

4 0
3 years ago
A quartz crystal vibrates with a frequency of 93621 Hz. What is the period of the crystals motion? T=1/93621 does not give me th
hram777 [196]

Answer:0.0000107s

Explanation:

f=1/T

T=1/f

f=93621 1/s

t=1/93621

T=0.00001068  s

3 0
3 years ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 n frictional force. he pushes in a dire
alexandr1967 [171]

Work is calculated as the product of Force, Distance, and angular motion. In this case, the work done by gravity is perpendicular to the motion of the cart, so θ = 90°

 and W=Fdcosθ

W=35.0 N x 20.0 m x cos90

W=0 J

This means that work done perpendicular to the direction of the motion is always zero.

5 0
3 years ago
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