Heated air moves from baseboard heaters to the rest of a room in a process called

Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first bl

o-na [289]

Answer:

The tension in the string connecting block 50 to block 51 is 50 N.

Explanation:

Given that,

Number of block = 100

Force = 100 N

let m be the mass of each block.

We need to calculate the net force acting on the 100th block

Using second law of newton

$F=ma$

$100=100m\times a$

$ma=1\ N$

We need to calculate the tension in the string between blocks 99 and 100

Using formula of force

$F_{100-99}=ma$

$F_{100-99}=1$

We need to calculate the total number of masses attached to the string

Using formula for mass

$m'=(100-50)m$

$m'=50m$

We need to calculate the tension in the string connecting block 50 to block 51

Using formula of tension

$F_{50}=m'a$

Put the value into the formula

$F_{50}=50m\times a$

$F_{50}=50\times1$

$F_{50}=50\ N$

Hence, The tension in the string connecting block 50 to block 51 is 50 N.

8 0

3 years ago

A car going 22 m/s increases its speed to pass a truck. Five seconds later the car is going 35 m/s. Calculate the acceleration o

jonny [76]

Answer:

I don't know

Explanation:

4 0

3 years ago

Read 2 more answers

A stone of mass m = 1.05 kg is released from a height of h = 2.1 m into a pool of water. At a time of t = 1.83 s after hitting t

mote1985 [20]

Answer:

Explanation:

ignoring air resistance, the kinetic energy at water impact will equal the potential energy converted

½mv² = mgh

v = √(2gh)

v = √(2(9.81)2.1) = 6.4188... m/s

after impact, an impulse will result in a change of momentum.

There is a downward impulse due to gravity equal to the weight of the stone and an upward average force due to water resistance and buoyancy force.

FΔt = mΔv

(F - mg)Δt = m(vf - vi)

(F - mg) = m(vf - vi)/Δt

F = m(vf - vi)/Δt + mg

F = m((vf - vi)/Δt + g)

F = 1.05(((½(-6.4188) - -6.4188)/ 1.83) + 9.81)

F = 12.14198...

F = 12.1 N

7 0

2 years ago

An empty 2,500 kg train car is headed northbound at a velocity of 5 m/s. Ahead of the first car, an empty 1,500 kg car is headed

Tema [17]

Let the mass of 2500 kg car be $m_1$ and it's velocity be $v_1$ and the mass of 1500 kg car be $m_2$ and it's velocity be $v_2$ .

After the bumping the mass be M and it's velocity be V.

By law of conservation of momentum we have

$m_1v_1+m_2v_2 = MV$

2500 * 5 + 1500 * 1=4000 * V

V = 14000/4000 = 7/2 = 3.5 m/s

So the velocity of the two-car train = 3.5 m/s

9 0

3 years ago

You want to slide a 0.39 kg book across a table. If the coefficient of kinetic friction is .21, what force is required to move t

uranmaximum [27]

Find the force that would be required in the absence of friction first, then calculate the force of friction and add them together. This is done because the friction force is going to have to be compensated for. We will need that much more force than we otherwise would to achieve the desired acceleration:

$F_{NoFric}=ma=0.39kg \times0.18 \frac{m}{s^2} =0.0702N$

The friction force will be given by the normal force times the coefficient of friction. Here the normal force is just its weight, mg

$F_{Fric}=0.39kg \times 9.8 \frac{m}{s^2} \times 0.21=0.803N$

Now the total force required is:

0.0702N+0.803N=0.873N

5 0

3 years ago