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saw5 [17]
3 years ago
14

Which describes what an outside observer at rest would observe about a spaceship that speeds up and approaches the speed of ligh

t?
Time moves faster on the spaceship, and the length of the spaceship decreases. Time moves faster on the spaceship, and the length of the spaceship increases. Time moves more slowly on the spaceship, and the length of the spaceship decreases.
Time moves more slowly on the spaceship, and the length of the spaceship increases.
Physics
2 answers:
Rzqust [24]3 years ago
7 0

Answer:

The correct answer choice is C.  

"Time moves more slowly on the spaceship, and the length of the spaceship decreases."

Explanation:

Answer on Q'let

Art [367]3 years ago
6 0
The outside observer, at rest relative to the spaceship, would see the spaceship
get shorter. and the clocks on the spaceship run slower than they should.

At the same time, the crew of the spaceship, looking back at the observer on
Earth, would see the observer on Earth get shorter, and the observer's clock
run slower than it should.

They would both be measuring what they see correctly.
You might be interested in
In the parts that follow select whether the number presented in statement A is greater than, less than, or equal to the number p
Brums [2.3K]

Answer:

a) Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

Explanation:

a) Statement A : 2.567km to two significant figures.

2.567km 2. S.F = 2.6km

Statement B : 2.567km to three significant figures.

2.567km 3 S.F = 2.57km

Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) statement A: (2.567 km + 3.146km) to 2 S.F

(2.567km + 3.146km) = 5.713km to 2 S.F = 5.7km

Statement B : (2.567 km, to two significant figures) + (3.146 km, to two significant figures).

2.567km to 2 S.F = 2.6km

3.146km to 2 S.F = 3.1km

2.6km + 3.1km = 5.7km

Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

5 0
2 years ago
Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4
Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0
3 years ago
Una locomotiva traina un solo vagone, che ha la sua stessa massa, con la forza di
gregori [183]

Answer:

c)    F = 16000 N

Explanation:

For this exercise we use Newton's second law

         F = ma

they tell us that adding the other wagons the acceleration of the locomotive must be maintained

 

      F = m a

by adding the other four wagons

mass = 4 no

therefore to maintain the force you must also raise the same factor

         Fe = 4Fo

         Fe = 4 4000

         F = 16000 N

3 0
3 years ago
1. An object on Earth and the same object on the Moon would have a difference in
Feliz [49]

Answers: (1) a. weight, (2)b. Force changes by 2/9, (3)b. movement, (4)a. 40,000 Joules, (5)c. the soil will be 5°C.

<h2>Answer 1: a. weight</h2>

Mass and weight are very different concepts.  

Mass is the amount of matter that exists in a body, which only depends on the quantity and type of particles within it. This means mass is an intrinsic property of each body and remains the same regardless of where the body is located.  

On the other hand, weight is a measure of the gravitational force acting on an object and is directly proportional to the product of the mass m of the body by the acceleration of gravity g:  

W=m.g  

Then, since the Earth and the Moon have different values ​​of gravity, t<u>he weight of an object in each place will vary</u>, but its mass will not.

<h2>Answer 2: b. Force changes by 2/9</h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:  

F=G\frac{m_{1}m_{2}}{r^2} (1)

Where:  

F is the module of the force exerted between both bodies  

G is the universal gravitation constant

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

If we double the mass of one object (for example 2m_{1}) and triple the distance between both (for example 3r). The equation (1) will be rewritten as:

F=G\frac{2m_{1}m_{2}}{(3r)^2} (2)

F=\frac{2}{9}G\frac{m_{1}m_{2}}{r^2} (3)

If we compare (1) and (2) we will be able to see the force changes by 2/9.

<h2>Answer 3: b. movement</h2>

The Work W done by a Force F refers to the release of potential energy from a body that is <u>moved</u> by the application of that force to overcome a resistance along a path.  

When the applied force is constant and <u>the direction of the force and the direction of the movement are parallel,</u> the equation to calculate it is:  

W=(F)(d)

Now, <u>when they are not parallel, both directions form an angle</u>, let's call it \alpha. In that case the expression to calculate the Work is:  

W=Fdcos{\alpha}

Therefore, pushing on a rock accomplishes no work unless there is movement (independently of the fact that movement is parallel to the applied force or not).

<h2>Answer 4: a. 40,000 Joules</h2>

The Kinetic Energy is given by:

K=\frac{1}{2}mV^{2}   (4)

Where m is the mass of the body and V its velocity

For the first case (kinetic energy K_{1}=10000J  for a car at V_{1}=30 mph=13.4112m/s):

K_{1}=\frac{1}{2}mV_{1}^{2}   (5)

Finding m:

m=\frac{2K_{1}}{V_{1}^{2}}   (6)

m=\frac{2(10000J)}{(13.4112m/s)^{2}}   (7)

m=111.197kg   (8)

For the second case (unknown kinetic energy K_{2}  for a car with the same mass at V_{2}=60 mph=26.8224m/s):

K_{2}=\frac{1}{2}mV_{2}^{2}   (9)

K_{2}=\frac{1}{2}(111.197kg)(26.8224m/s)^{2}   (10)

K_{2}=40000J   (11)

<h2>Answer 5: c. the soil will be 5°C</h2>

The formula to calculate the amount of calories Q is:

Q=m. c. \Delta T   (12)

Where:

m  is the mass

c  is the specific heat of the element. For water is c_{w}=1 kcal/g\°C  and for soil is c_{s}=0.20 kcal/g\°C  

\Delta T  is the variation in temperature (the amount we want to find for both elements)

This means we have to clear \Delta T from (12) :

\Delta T=\frac{Q}{m.c}   (13)

For Water:

\Delta T_{w}=\frac{Q_{w}}{m_{w}.c_{w}}   (14)

\Delta T_{w}=\frac{1kcal}{(1kg)(1 kcal/g\°C)}   (15)

\Delta T_{w}=1\°C)}   (16)

For Soil:

\Delta T_{s}=\frac{Q_{s}}{m_{s.c_{s}}   (17)

\Delta T_{s}=\frac{1kcal}{(1kg)(0.20 kcal/g\°C)}   (18)

\Delta T_{s}=5\°C)}   (19)

Hence the correct option is c.

5 0
3 years ago
Read 2 more answers
A curve of radius 53.1 m is banked so that a car of mass 2.9 Mg traveling with uniform speed 67 km/hr can round the curve withou
prisoha [69]

Answer:

33.65°

Explanation:

radius, r = 53.1 m

m = 2.9 Mg = 2.9 x 10^6 g = 2900 kg

v = 67 km/h

convert km/h into m/s

v = 18.61 m/s

Let the angle of banking of road is θ, without friction

tan\theta =\frac{v^{2}}{rg}

tan\theta =\frac{18.61^{2}}{53.1\times 9.8}

tan  θ = 0.6655

θ = 33.65°

Thus, the angle of banking of road is 33.65°.

6 0
3 years ago
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