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Tamiku [17]
4 years ago
7

How many joules of energy are required to accelerate a 1.0-kg mass from rest to a speed of 86.6% the speed of light?

Physics
1 answer:
deff fn [24]4 years ago
4 0
KE = ½m*v² = ½*1.0*[0.866*3E8]² = 3.375E16 J 

<span>Etot = mc²/√[1 - (v/c)²] = 1.8E17 J</span>
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what is the density of a substance that has a mass of 2.0 g , and when placed in a graduated cylinder the volume changed from 70
lubasha [3.4K]

A material with a mass of 2.0 g when placed in a graduated cylinder the volume changed from 70 ml to 75 ml has a density of 0.4 g/mL.

How do I calculate the substance's density?

We'll start by getting the substance's volume. This is attainable as follows:

Water volume: 70 mL

75 mL = volume of material + water.

Substance volume =?

Substance volume equals (substance volume plus water) - (Volume of water)

Substance volume = 75 - 70

5 mL is the substance's volume.

Finally, we will calculate the substance's density. Below is an example to help:

2.0 g is the substance's mass.

5 mL is the substance's volume.

Substance density =?

Mass / volume equals density.

Substance density = 2/5

0.4 g/mL is the substance's density.

The density is therefore 0.4 g/mL.

To know more about density, visit:

brainly.com/question/952755

#SPJ4

6 0
1 year ago
A block of weight mg sits on an inclined plane as shown in (Figure 1) . A force of magnitude F1 is applied to pull the block up
svlad2 [7]

Answer:

W = (F1 - mg sin θ) L,   W = -μ  mg cos θ L

Explanation:

Let's use Newton's second law to find the friction force. In these problems the x axis is taken parallel to the plane and the y axis perpendicular to the plane

Y Axis  

       N - W_{y} =

       N = W_{y}

X axis

       F1 - fr - Wₓ = 0

       fr = F1 - Wₓ

Let's use trigonometry to find the components of the weight

     sin θ = Wₓ / W

     cos θ = W_{y} / W

      Wₓ = W sin θ

      W_{y} = W cos θ

We substitute

      fr = F1 - W sin θ

Work is defined by

        W = F .dx

        W = F dx cos θ

The friction force is parallel to the plane in the negative direction and the displacement is positive along the plane, so the Angle is 180º and the cos θ= -1

         

        W = -fr x

        W = (F1 - mg sin θ) L

Another way to calculate is

         fr = μ N

         fr = μ W cos θ

the work is

         W = -μ  mg cos θ L

4 0
3 years ago
Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the
Doss [256]

Answer:

v_{su} = 19.44 m/s

Explanation:

m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg

T=9.29x10^8\\r_{o}=1.43x10^{12}

If the sun considered as x=0 on the axis to put the center of the mass as a:

m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}

solve to r1

r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}

r_1=1.428x10^9m

Now convert to coordinates centered on the center of mass.  call the new coordinates x' and y' (we won't need y').  Now since in the sun centered coordinates the angular momentum was  

L = \frac{m_{sa}*2*pi*r_1^2}{T}

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum.  So that means

L_{sun}=\frac{m_{sa}*2*\pi *( 2r_{o}*r_1 -r_1^2)}{T}

Since

L_{su}= m_{su}*v_{su}*r_1

then

v_{su}=\frac{m_{sa}*2*pi*(2r_{o}*r_{1}-r_{1}^2)}{T*m_{sa}*r_1}

v_{su} = 19.44 m/s

7 0
4 years ago
If the tension of the cable is 25.0 N what is the mass of the ball
marysya [2.9K]

Answer:576

Explanation:

5 0
3 years ago
A bug splats against the windshield of a car traveling at high speeds down a backcountry road. Which statement correctly compare
zvonat [6]

Answer:

C. The bug's change in momentum is equal to the car's change in momentum.

Explanation:

As we know by Newton's 2nd law

F = \frac{\Delta P}{\Delta t}

here we have also know that when car hits the bug then force applied by wind shield on the bug is same as the force applied by the bug on the car's wind shield as per Newton's III law

F_{12} = F_{21}

so we know that

\frac{\Delta P_{12}}{\Delta t} = \frac{\Delta P_{21}}{\Delta t}

so we have

\Delta P_{12} = \Delta P_{21}

so correct answer will be

C. The bug's change in momentum is equal to the car's change in momentum.

6 0
4 years ago
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