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ki77a [65]
3 years ago
15

Two steel blocks rest on a frictionless, horizontal surface. Block X has a mass of

Physics
1 answer:
stepan [7]3 years ago
3 0
Before you jump into this, you have to stand back, look at it, and decide what's going to happen.

You're pulling on some things that rest on a frictionless surface, so they have to start moving.  In fact, they're going to accelerate.

Now, I want you to consider for just a second:  What would happen if
the two blocks were glued together ?  Then, you'd be pulling on a single
18-kg block, with 36N of force, and it would accelerate to the right at
[ A = F/M ] = 36/18 = 2 m/s² .

The actual situation is no different.  The string between them has no weight,
and it's tight, and there are no other forces on the blocks, so the whole combination ... both blocks ... is still going to accelerate to the right at 2 m/s² .

OK.  Now, let's look at the individual blocks:

The one in the back ... the 12 kg.  It's accelerating at 2 m/s² to the right,
so the force on it must be  [ F = M A ] = 12 x 2 = 24 N to the right.  Where
does that force come from ?  It's the tension in the string between them. 

The one in the front ... the 6-kg.
It has 36 N pulling it to the right, and 24 N of string tension pulling it to the left.
The net force on this block is  (26 - 24) = 12N to the right.
Its acceleration is  [ A = F/M ] = 12/6 = 2m/s² to the right.

Well, whaddaya know !  Both blocks accelerate at the same rate, in the
same direction, just as if they were glued together, and the string between
them remains tight, with 24N of tension in it.

With this description, you can handle the FBD on your own, I'm sure.
You might be interested in
The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee
kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2>v_f = 35.3 m/s</h2>

Explanation:

Part a)

As we know that drag force is given as

F = \frac{C_d \rho A v^2}{2}

C_d = 0.35

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

A = \frac{\pi(0.074)^2}{4}

A = 4.3 \times 10^{-3} m^2

v = 40.2 m/s

so we have

F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}

F = 1.46 N

So acceleration of the ball is

a = \frac{F}{m}

a = \frac{1.46}{0.145}

a = 10.1 m/s^2

Part B)

As per kinematics we know that

v_f^2 - v_i^2 = 2 a d

v_f^2 - 40.2^2 = 2(-10.1)(18.4)

v_f = 35.3 m/s

4 0
3 years ago
Tarik winds a small paper tube uniformly with 183 turns 183 turns of thin wire to form a solenoid. The tube's diameter is 9.49 m
Rufina [12.5K]
<h2>Answer:</h2>

143μH

<h2>Explanation:</h2>

The inductance (L) of a coil wire (e.g solenoid) is given by;

L = μ₀N²A / l                 --------------(i)

Where;

l = the length of the solenoid

A = cross-sectional area of the solenoid

N= number of turns of the solenoid

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

<em>From the question;</em>

N = 183 turns

l = 2.09cm = 0.0209m

diameter, d = 9.49mm = 0.00949m

<em>But;</em>

A = π d² / 4                     [Take π = 3.142 and substitute d = 0.00949m]

A = 3.142 x 0.00949² / 4

A = 7.1 x 10⁻⁵m²

<em>Substitute these values into equation (i) as follows;</em>

L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209           [Take π = 3.142]

L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209

L = 143 x 10⁻⁶ H

L = 143 μH

Therefore the inductance in microhenrys of the Tarik's solenoid is 143

6 0
3 years ago
2) How could a motorcycle (a vehicle with less mass) make a van (a
jeyben [28]

it depends upon what state they are in like in motion or res

8 0
2 years ago
A beam of white light enters a prism of crown glass (n = 1.5) from air (n = 1.00). Once inside, the colors in the light disperse
eduard

Answer:

= 2.33

Explanation:

.According to snell's law:

n1sin i = n2sin r ,

where n1 is refractive index of the medium in which incident ray is travelling, n2 is the refractive index of the medium in which refracted ray is travelling,

i is angle of incidence,

r is angle of refraction.

Given that,

n1 = 1,

i = 51 degrees,

r = 19.5 degrees. ,

n2= ?

So,

1*sin 51 = n2 sin 19.5  

=> n2 = sin51 / sin19.5

 = 2.33

7 0
3 years ago
A meter stick moves parallel to its axis with speed of 0.96 c relative to you. What would you measure for the length of the stic
Fed [463]

Answer:

The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

Explanation:

Given that,

Relative speed of stick v= 0.96 c

Speed of light c= 2.99793\times10^{8}\ m/s

Proper length of stick = 1 m

We need to calculate the length of the stick

Using formula of length

\Delta l=\Delta l_{0}\sqrt{(1-\dfrac{v^2}{c^2})}

Put the value into the formula

\Delta l=1\sqrt{1-\dfrac{(0.96)^2c^2}{c^2}}

\Delta l=1\sqrt{1-(0.96)^2}

\Delta l=0.28\ m

We need to calculate the time the stick take to move

Using formula of time

t=\dfrac{\Delta l}{v}

Put the value into the formula

t=\dfrac{0.28}{0.96\times(2.99793\times10^{8})}

t=9.72\times10^{-10}\ sec

t=0.97\ ns

Hence, The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

7 0
3 years ago
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