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3 years ago

How to find instantaneous velocity physics?

1 answer:

8 0

You should have the velocity as a function of time either given explicitly or implicitly (a graph)

v = ds/dt (differentiating the position vector)

integrating the acceleration.

you can use impulse or work and energy principle and also newton law of motion to find acceleration then velocity

NOT SURE IF THAT WHAT YOU WANT.

You might be interested in

A woman pulls on a 6.00-kg crate, which in turn is connected to a 4.00-kg

djverab [1.8K]

Answer:

B. is subject to a smaller net force but same acceleration.

Explanation:

F = m*a

So because our force applied is constant from the women pulling on the rope which means the acceleration is the same on both the 4kg create and the 6kg create. The only thing that changes here is the mass of the creates, so there is more tension force between the women and the 6kg create then there is between the 4kg create and the 6kg. It takes less force to move the 4kg create therefore the tension force is less between the two creates.

7 0

3 years ago

Read 2 more answers

One of the dangers of tornados and hurricanes is the rapid drop in air pressurethat is associated with such storms. Assume that

Oksi-84 [34.3K]

Answer:

The net outward force is exerted on a square window of the house is $4.5930\times10^{4}\ N$ .

Explanation:

Given that,

Pressure = 1.02 atm

External air pressure = 0.910 atm

Distance = 2.03 m

We need to calculate the net outward force

Using formula of force

$F= PA$

Where, P = pressure

F = force

A = area

Put the value into the formula

$F=(1.02-0.910)\times101325\times(2.03)^2$

$F=4.5930\times10^{4}\ N$

Hence, The net outward force is exerted on a square window of the house is $4.5930\times10^{4}\ N$ .

6 0

4 years ago

Two identical blocks are attached to the same massless rope, which is strung around two massless, frictionless pulleys. A massle

Alik [6]

Answer:

The value of mass 1, m1= 6/5m

The value of mass 2, m2= 3/5m

Explanation:

case 1:

here tension and the acceleration will be:

for m1;

mg-T=ma
2mg - 2T = 2ma .....1.

for m2:

2T-mg = ma/2 ..... 2.

adding the both equations,

2mg - 2T + 2T-mg = 2ma + ma/2

a = 2/5 g

putting the value of a into the equation 1.

mg - T = m* (2/5)g

T = 3/5 ( mg )

now

case 2:

The two identical blocks are released from the rest, the tension remains the same as the case 1.

so,

for m1:

2T-m2g=0

for m2:

2m2g - 2T =0

adding both equations we get,

2T-m2g + 2m2g - 2T = 0

m2 = m1 / 2

T = m1*g / 2

here we know that

T (case1) = T (case2)

3/5 ( mg ) = m1*g / 2

m1 = 6/5 m

hence

m2 = 3/5 m

learn more about tension here:

<u>brainly.com/question/23590078</u>

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#SPJ4

4 0

2 years ago

Explain the melting of a solid in terms of molecules and energy

Pie

When you melt a solid the heat creates energy, when you heat up molecules they will move quicker and start to somewhat break apart into a liquid structure.

4 0

3 years ago

A bullet with mass 0.04000 kg, traveling at a speed of 310.0 m/s, strikes a stationary block of wood having mass 6.960 kg. The b

ch4aika [34]

Answer:

1.77 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m')............................ Equation 1

Where m = mass of the bullet, m' = mass of the block, u = initial velocity of the bullet, u' = initial velocity of the block, V = Velocity of the bullet-wood combination immediately after collision

make V the subject of the equation

V = (mu+m'u')/(m+m').................. Equation 2

Given: m = 0.04 kg, m' = 6.96 kg, u = 310 m/s, u' = 0 m/s (stationary)

Substitute into equation 2

V = (0.04×310+6.96×0)/(0.04+6.96)

V = 12.4/7

V = 1.77 m/s

Hence the speed, in m/s, of the bullet-plus-wood combination immediately after the collision is 1.77 m/s

6 0

4 years ago