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MA_775_DIABLO [31]
1 year ago
12

A rod with a rest length of 1m whizzes past an observer at a speed of 0.995c, where c is the speed of light. Is it possible that

the observer could also measure the length of the moving rod to be 1 m? A. yes B. no
Physics
2 answers:
Tresset [83]1 year ago
6 0

Answer:

B. no

Explanation:

  • When any body moves at a speed comparable to the speed of light (<em>i.e. relativistic speed</em>) then the observer sees a contraction in length of the body along the axis of motion.

Assuming that the motion of the body is along the axis of the rod, then the observer will measure its length to be lesser than its length at rest.

<u>Then according to Einstein's theory of relativity:</u>

L=\frac{L_0}{\gamma}

where:

L_0=  original length of the object (along the direction of motion)

L = observed length of the rod

\gamma = Lorentz factor

\gamma\equiv \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }

abruzzese [7]1 year ago
6 0

Answer:

B. no

Explanation:

If any body moves at a rate equal to the speed of light (i.e. relativistic velocity), the observer experiences a contraction along the direction of motion in the span of the body.

Lets us suppose that the motion of the body is along the axis of the rod, the observer will measure the length of the rod which will be lesser than its length in rest position.

And contracted length is given by

L=L_o\sqrt{1-\frac{v^2}{c^2} }

Lo= original length of the object along its axis.

L= length measured by the observer.

v= 0.995c

c= speed of light

So,observer will measure a length less than 1 m

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20 points please help!!
IgorC [24]

Answer:

Sample Response: If temperature and surface area increase, then the time it takes for sodium bicarbonate to completely dissolve will decrease, because increasing both factors increases the rate of a chemical reaction.

Explanation:

4 0
2 years ago
What are the magnitude and direction of the force the pitcher exerts on the ball? (enter your magnitude to at least one decimal
murzikaleks [220]
Details are missing in the question. Complete text of the problem:

"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.

(a) Through what distance does it move before its release? (m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball? (Enter your magnitude to at least one decimal place.)"


Solution

(a) The pitcher accelerates the baseball from rest to a final velocity of v_f = 16.5 m/s, so \Delta v=16.5 m/s, in a time interval of \Delta t = 181 ms=0.181 s. The acceleration of the ball in the horizontal direction (x-axis) is therefore

a_x =  \frac{\Delta v}{\Delta t}= \frac{16.5 m/s}{0.181 s}=91.2 m/s^2

And the distance covered by the ball during this time interval, before it is released, is:

S= \frac{1}{2} a_x (\Delta t)^2 = \frac{1}{2} (91.2 m/s^2)(0.181 s)^2=1.49 m

(b) For this part we need to consider also the weight of the ball, which is W=mg=2.28 N

From this, we find its mass: m= \frac{W}{g}= \frac{2.28 N}{9.81 m/s^2}=0.23 Kg

Now we can calculate the magnitude of the force the pitcher exerts on the ball. On the x-axis, we have

F_x = m a_x = (0.23 kg)(91.2 m/s^2)=20.98 N

We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:

F_y=W=mg=2.28 N (upward)

So, the magnitude of the force is

F= \sqrt{F_x^2+F_y^2}=  \sqrt{(20.98N)^2+(2.28N)^2}=21.2 N

To find the direction, we should find the angle of F with respect to the horizontal. This is given by

\tan \alpha =  \frac{F_y}{F_x}= \frac{2.28 N}{20.98 N}=0.11

From which we find \alpha=6.2^{\circ}

7 0
2 years ago
Read 2 more answers
A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c
Mrac [35]

Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Radius = 0.15 mm

Current densityJ=2.8\times10^{7}\ A/m^2

We need to calculate the current

Using formula of current density

J = \dfrac{I}{A}

I=J\timesA

Where, J = current density

A = area

I = current

Put the value into the formula

I=2.8\times10^{7}\times\pi\times(0.15\times10^{-3})^2

I=1.97=2.0\ A

Hence, The current is 2.0 A.

7 0
2 years ago
A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'
lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3  

                                              = 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

Y =\frac{mgL}{A\Delta L}

\Delta L =\frac{mgL}{A\times Y}

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0
2 years ago
Io, a satellite of Jupiter, is the most volcanically active moon or planet in the solar system. It has volcanoes that send plume
Mamont248 [21]

Answer:

1331.84 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement = 490 km

a = Acceleration

g = Acceleration due to gravity = 1.81 m/s² = a

From equation of linear motion

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -1.81\times 490000}\\\Rightarrow u=1331.84\ m/s

The speed of the material must be 1331.84 m/s in order to reach the height of 490 km

3 0
2 years ago
Read 2 more answers
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