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pantera1 [17]
2 years ago
5

1. The gravitational pull of the sun on Earth keeps Earth orbiting around the sun. Which statement is correct about the force th

at Earth exerts on the sun?(1 point)
Earth pulls the sun away from itself with a force equal to the ratio of the mass of the sun to the mass of Earth.

Earth pulls the sun away from itself with an equal force.

Earth pulls the sun toward itself with an equal force.

Earth pulls the sun toward itself with a force equal to the ratio of the mass of the sun to the mass of Earth.

2. Two rockets with the same mass are accelerated. Rocket A accelerates twice as quickly as rocket B. Which statement is correct?(1 point)

The motor in rocket A is twice as powerful as the motor in rocket B.

The motor in rocket A is four times as powerful as the motor in rocket B.

The motor in rocket A is half as powerful as the motor in rocket B.

The motor in rocket A is half as powerful as the motor in rocket B.

3. Five motorboats are being tested to see which reaches the highest velocity in the same amount of time. After graphing the acceleration versus force for each motorboat, the graph is a sloped line with a y-intercept of zero. Which statement is correct about these motorboats?(1 point)

All five motorboats have the same mass, which cannot be calculated from the graph.

All five motorboats have the same final velocity, which cannot be calculated from the graph.

All five motorboats have the same final velocity, which can be calculated from the graph.

All five motorboats have the same mass, which can be calculated from the graph.

4. Three objects are released to the ground while the mass and net force of each is recorded. Air resistance is assumed to be negligible. Which statement is correct about all three objects?(1 point)

No statements can be made about the objects without measuring the acceleration of each.

The ratio of mass to acceleration is constant across all objects.

The ratio of net force to acceleration is constant across all objects.

The ratio of net force to mass is constant across all objects.

5. A helicopter is accelerating up into the air while it is also accelerating due north. Which statement about the forces acting on the helicopter is true?(1 point)

The drag on the helicopter is directed downward and to the south.

The drag on the helicopter is directed upward and to the south.

The drag on the helicopter is directed downward and to the north.

The drag on the helicopter is directed upward and to the north.
Physics
2 answers:
Ludmilka [50]2 years ago
6 0

Answer:

1. Earth pulls the sun away from itself with an equal force.

Mice21 [21]2 years ago
6 0

Answer:

Rocket A is twice as powerful

You might be interested in
Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0
2 years ago
A basketball with mass of 0.8 kg is moving to the right with velocity 6 m/s and hits a volleyball with mass of 0.6 kg that stays
IceJOKER [234]

Answer:

26.67 m/s

Explanation:

From the law of conservation of linear momentum, the initial sum of momentum equals the final sum.

p=mv where p is momentum, m is the mass of object and v is the speed of the object

Initial momentum

The initial momentum will be that of basketball and volleyball, Since basketball is initially at rest, its initial velocity is zero

p_i= m_bv_b+m_vv_v=8*6+0.6*0=48 Kg.m/s

Final momentum

p_f= m_bv_b+m_vv_v=8*4+0.6*v_v=32+0.6v Kg.m/s\\32+0.6v_v=48\\0.6v=16\\v_v=16/0.6=26.66666667\approx 26.67 m/s

4 0
2 years ago
Electromagnetic radiation is emitted when a charged particle moves through a medium faster than the local speed of light. This r
alexandr1967 [171]

Answer:

to create the particle the speed must be greater than 2.25 10⁸ m / s

Explanation:

In this exercise we must use the relation of the index of refraction with the speed of light in a vacuum and a material medium

           n = c / v

where c is the speed of light in the vacuum, v the speed of light in the material medium and n the ratio of rafraccio

in this case they give us that the medium matter water them that has a refractive index of

              n = 1,333

we clear

          v = c / n

let's calculate

           v = 3 10⁸ / 1,333

           v = 2.25 10⁸ m / s

to create the particle the speed must be greater than 2.25 10⁸ m / s

6 0
2 years ago
The leaning tower of Pisa is about 56 meters tall. A ball released from the top takes 3.4 seconds to reach the ground. The final
geniusboy [140]
S=56, u=0, v=33, a=?, t=3.4

v=u+at
33=3.4 a
a = 9.7m/s^2
7 0
2 years ago
Read 2 more answers
ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on it
Nadya [2.5K]

Answer:

  K_b = 78 J

Explanation:

For this exercise we can use the conservation of energy relations

starting point. Lowest of the trajectory

        Em₀ = K = ½ mv²

final point. When it is at tea = 50º

        Em_f = K + U

        Em_f = ½ m v_b² + m g h

where h is the height from the lowest point

        h = L - L cos 50

        Em_f = ½ m v_b² + mg L (1 - cos50)

energy be conserve

        Em₀ = Em_f

         ½ mv² = ½ m v_b² + mg L (1 - cos50)

         K_b = ½ m v_b² + mg L (1 - cos50)

let's calculate

          K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)

          K_b = 36 +42.0

          K_b = 78 J

4 0
2 years ago
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