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yanalaym [24]
3 years ago
8

Light rays that enter the hole of a pinhole camera form _____ on the back wall of the camera.

Physics
1 answer:
11111nata11111 [884]3 years ago
5 0
That's an inverted real image.
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What is the net force on an object that has a force pushing downward at 25N and a forcing pushing upward at 10N?
shutvik [7]

Answer:

15N downwards

Explanation:

Net force is the sum of all forces acting on an object.

Taking upwards as positive and downwards as negative,

Net Force = 10N + (-25N)

= -15N

Hence <u>Net Force is 15N downwards</u>.

7 0
4 years ago
Read 2 more answers
Help me please please help me
andreev551 [17]
Answers:

1. B. Circuit
2. D. Electrons moving around
3. B. Series and parallel
4. C. Parallel
5. C. Series
6. A. One
7. C. If one goes down, they both go down.

Hope this helped!
5 0
3 years ago
A truck covers 40.0 m in 7.80 s while uniformly slowing down to a final velocity of 1.70 m/s. Please provide with explanation an
Karolina [17]

Answer:

a = -0.223 m/s

Explanation:

4 0
3 years ago
Commercially-available hybrid vehicles, such as the Toyota Prius, use electrical batteries to store energy for later use. Howeve
Bad White [126]

(A) 4.2\cdot 10^5 J

The energy stored by the system is given by

E=Pt

where

P is the power provided

t is the time elapsed

In this case, we have

P = 60 kW = 60,000 W is the power

t = 7 is the time

Therefore, the energy stored by the system is

E=(60,000 W)(7 s)=4.2\cdot 10^5 J

(B) 4830 rad/s

The rotational energy of the wheel is given by

E=\frac{1}{2}I \omega^2 (1)

where

I is the moment of inertia

\omega is the angular velocity

The moment of inertia of the wheel is

I=\frac{1}{2}MR^2=\frac{1}{2}(5 kg)(0.12 m)^2=0.036 kg m^2

where M is the mass and R the radius of the wheel.

We also know that the energy provided is

E=4.2\cdot 10^5 J

So we can rearrange eq.(1) to find the angular velocity:

\omega=\sqrt{\frac{2E}{I}}=\sqrt{\frac{2(4.2\cdot 10^5 J)}{0.036 kg m^2}}=4830 rad/s

(C) 2.8\cdot 10^6 m/s^2

The centripetal acceleration of a point on the edge is given by

a=\omega^2 R

where

\omega=4830 rad/s is the angular velocity

R = 0.12 m is the radius of the wheel

Substituting, we find

a=(4830 rad/s)^2 (0.12 m)=2.8\cdot 10^6 m/s^2

7 0
4 years ago
A point charge Q at the center of a sphere of radius R produces an electric flux of Φ coming out of the sphere. If the charge r
Dafna1 [17]

Remains the same

Explanation:

According to Gauss's law, the electric flux through a closed surface is proportional to the charge enclosed by the surface. So no matter how big or small we make the surface that encloses the charge, the electric flux remains the same because it only depends on the enclosed charge, not surface area.

3 0
3 years ago
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