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OleMash [197]
3 years ago
13

PLEASE HELP ME WITH THIS ONE QUESTION

Physics
1 answer:
allochka39001 [22]3 years ago
5 0

Answer:

the image will be real and inverted

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Based on Kepler's work, which best describes the orbit If a planet around the Sun?
krek1111 [17]

Answer:

an ellipse with the Sun at one focus  or D

Explanation:

Answer for edgenuity

7 0
3 years ago
Two blocks, A and B, are connected by a rope. A second rope is connected to block B and a steady, horizontal tension force of T
Oksanka [162]

Answer:

40 N

Explanation:

We are given that

Speed of system is constant

Therefore, acceleration=a=0

Tension applied on block B=T=50 N

Friction force=f=10 N

We have to find the friction force acting on block A.

Let T' be the tension in string connecting block A and block B and friction force on block A be f'.

For Block B

T-f-T'=m_Ba

Where m_B=Mass of block B

Substitute the values

50-10-T'=m_B\times 0=0

T'==40 N

For block A

T'-f'=m_Aa

Where m_A=Mass of block A

Substitute the values

40-f'=m_A\times 0=0

f'=40 N

Hence, the friction force acting on block A=40 N

3 0
3 years ago
What travels faster than anything anything else in the universe?
cluponka [151]

Answer:

The announcement he had made promised to overturn our understanding of the Universe. If the data gathered by 160 scientists working on the project were correct, the unthinkable had been observed. Particles – in this case, neutrinos – had travelled faster than light.

Explanation:  Plz Mark brainleist

4 0
2 years ago
A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro
svetoff [14.1K]

Answer:

The equation of motion is x(t)=-\frac{1}{3} cos4\sqrt{6t}

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is 32ft/s^2

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet x=\frac{4}{12} =\frac{1}{3}feet

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

W=kx ⇒ k=\frac{W}{x}

The spring constant , k=\frac{24}{1/3}

                            = 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

    m\frac{d^2x}{dt} +kx=0

    \frac{3}{4} \frac{d^2x}{dt} +72x=0

  \frac{d^2x}{dt} +96x=0

Auxiliary equation is, m^2+96=0

                                 m=\sqrt{-96}

                               =\frac{+}{} i4\sqrt{6}

Thus , the solution is x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}

                                 x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2  4\sqrt{6} cos4\sqrt{6t}

The mass is released from the rest x'(0) = 0

                    =-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2 4\sqrt{6} cos4\sqrt{6(0)} =0

                                                    c_2 4\sqrt{6} =0

                                     c_2=0

Therefore , x(t)=c_1 cos 4\sqrt{6t}

Since , the mass is released from the rest from 4 inches

                    x(0)= -4 inches

c_1 cos 4\sqrt{6(0)} =-\frac{4}{12} feet

   c_1=-\frac{1}{3} feet

Therefore , the equation of motion is  -\frac{1}{3} cos4\sqrt{6t}

7 0
3 years ago
An airplane starts from rest and reaches a takeoff velocity of 60.0m/s due north in 4.0 s. How far has it traveled before takeof
Blizzard [7]
240 meters that's it
7 0
3 years ago
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