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ratelena [41]
3 years ago
11

A motorcycle drives up a steeply inclined ramp. The work done on the motorcycle by the Earth’s gravitational force is ____. 1.

positive 2. negative 3. zero
Physics
1 answer:
Leto [7]3 years ago
8 0

Answer:

be

Explanation:

You might be interested in
It is proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of foil. W
Anvisha [2.4K]

Answer:

962291.57928 m²

Explanation:

P_r = Pressure = 2\dfrac{I}{c}  (full reflection)

I = Intensity = \dfrac{P}{A}=\dfrac{P}{4\pi r^2}

P = Power = 3.9\times 10^{26}\ W

c = Speed of light = 3\times 10^8\ m/s

M = Mass of Sun = 1.99\times 10^{30}\ kg

m = Mass of ship = 1500 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Force of radiation is given by

F_r=P_rA\\\Rightarrow F_r=2\dfrac{I}{c}\times A\\\Rightarrow F_r=2\dfrac{P}{4\pi r^2c} A

This force will balance the gravitational force as stated in the question

\dfrac{GMm}{r^2}=2\dfrac{P}{4\pi r^2c} A\\\Rightarrow A=\dfrac{4\pi cGMm}{2P}\\\Rightarrow A=\dfrac{4\times \pi\times 3\times 10^8\times 6.67\times 10^{-11}\times 1.99\times 10^{30}\times 1500}{2\times 3.9\times 10^{26}}\\\Rightarrow A=962291.57928\ m^2

The area of the must be 962291.57928 m²

3 0
3 years ago
A point charge is placed at each corner of square with side leanth a. The charges all have same magnitude q. My question, What i
nexus9112 [7]

Answer:

 E = k q / a²   (1.3535) (- i ^ + j ^)

  E = k q / a²  1.914  ,      θ’= 135

Explanation:

For this exercise we will use Newton's second law where we must add as vectors

        E_total = E₁₂ i ^ + E₁₄ j ^ + E₁₃

Let's look for the value of each term

On the x axis

       E₁₂ = k q / a²

On the y axis

       E₁₄ = k q / a²

For the charge in the opposite corner we look for the distance

        d = √ (a² + a²) = a √2

let's look for the field

      E₁₃ = k q / d²

      E₁₃ = k q / 2a²

let's use trigonometry to find the two components of this field

       cos 45 = E₁₃ₓ / E₁₃

       E₁₃ₓ = E₁₃ cos 45

       

       sin 45 = E_{13y} / E₁₃

       E_{13y} = E₁₃ sin 45

       E₁₃ₓ = k q / 2a²  cos 45

       E_{13y} = k q / 2a²  sin 45

let's find each component of the electric field

X axis

      Eₓ = -E₁₂ - E₁₃ₓ

      Eₓ = - k q / a² - k q / 2a² cos 45

      Eₓ = - k q / a² (1 + cos 45/2)

      cos 45 = sin 45 = 0.707

      Eₓ = - k q / a²   (1 + 0.707 / 2)

      Eₓ = - k q / a²    (1.3535)

Y axis  

      E_y = E₁₄ + E_{13y}

       E_y = k q / a² + k q / 2a²     sin 45

       E_y = k q / a² (1 + sin 45/2)

       E_y = k q / a²       (1.3535)

we can give the results in two ways

       E = k q / a²   (1.3535) (- i ^ + j ^)

In modulus and angle form, let's use Pythagoras' theorem for the angle

       E = √ (Eₓ² + E_y²)

        E = k q / a²    1.3535 √2

        E = k q / a²     1.914

we use trigonometry for the angle

        tan θ = E_y / Eₓ

         θ = tan⁻¹  (E_y / Eₓ)

         θ = tan⁻¹ (1 / -1)

         θ = 45

in the third quadrant, if we measure the angle of the positive side of the x-axis

           θ‘= 90 + 45

           θ’= 135

4 0
4 years ago
Two tugboats are moving a barge. Tugboat A pushes on the barge with a force of 3000n. Tugboat B pulls with a force of 5000 Newto
kenny6666 [7]

The net force on the barge is 8000 N

Explanation:

In order to find the net force on the badge, we have to use the rules of vector addition, since force is a vector quantity.

In this problem, we have two forces:

  • The force of tugboat A, F_A = 3000 N, acting in a certain direction
  • The force of tugboat B, F_B = 5000 N, also acting in the same direction

Since the two forces act in the same direction, this means that we can simply add their magnitudes to find the net combined force on the barge. Therefore, we get

F=F_A+F_B = 3000 + 5000 = 8000 N

and the direction is the same as the direction of the two forces.

Learn more about forces:

brainly.com/question/11179347

brainly.com/question/6268248

#LearnwithBrainly

5 0
4 years ago
In some cases fixture wires may be used for
zalisa [80]

You can use fixture wires: For installation in luminaires where they are enclosed and protected and not subject to bending and twisting and also can be used to connect luminaires to their branch circuit conductors.

<h3>What are some uses of fixture wires?</h3>

Fixture wires are flexible conductors that are used for wiring fixtures and control circuits. There are some special uses and requirements for fixture wires and no fixture can be smaller than 18 AWG

In modern fixtures, neutral wire is white and the hot wire is red or black. In some types of fixtures, both wires will be of the same color.

To know more about fixture wires, refer

brainly.com/question/26098282

#SPJ4

3 0
1 year ago
A 10 kg blue cart moving to the right at 25 m/s collides with a 17 kg red cart moving in the opposite direction at 16 m/s. If, a
scoray [572]

Answer:

24.8m/s

Explanation:

Given data

m1= 10kg

u1=25m/s

m2=17kg

u2=16m/s

v1=10m/s

v2=??

Applying the conservation of linear momentum

m1u1+m2u2=m1v1+m2v2

substitute

10*25+17*16=10*10+17*v2

250+272=100+17v2

522=100+17v2

522-100=17v2

422=17v2

Divide both sides by 17

v2= 422/17

v2= 24.8 m/s

Hence the velocity of the red cart is 24.8m/s in the opposite direction of the blue cart

3 0
3 years ago
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