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lawyer [7]
2 years ago
14

A car is moving north on a freeway. If a bug is flying south on the freeway, is the momentum of the bug positive or negative?

Physics
1 answer:
faltersainse [42]2 years ago
3 0

Negative

Because the car is moving up and the bug is moving down. but it also depends on the weather so choice between one of those two I think is Negative but I may be wrong.

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O be useful in science, a hypothesis must be *
shepuryov [24]

Answer:

the answer is c testable

3 0
3 years ago
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A ball is thrown upward with an initial velocity of 13 m/s. Using the approximate value of
dem82 [27]

Answer:

v=v0 - gt

Explanation:

The equation for velocity is

v=v0 - gt

where v0=14m/s, g=10m/s^2.

in 1 second:

v=14-10=4m/s

it is positive so direction is upwards

in 2 seconds:

v=14-20=-6m/s

it is negative so direction is downwards

5 0
2 years ago
The magnitude of each force is 208 N the force on the right is applied at an angle 36° and the mass of the block is 17 kg the co
djyliett [7]

Answer:

<em>11.06m/s²</em>

Explanation:

According to Newtons second law of motion

\sm F_x = ma_x\\F_m - F_f = ma_x\\mgsin \theta - \mu R mgcos \theta = ma_x\\

Given

Mass m = 17kg

Fm = 208N

theta = 36 degrees

g = 9.8m/s²

a is the acceleration

Substitute

208 - 0.148(17)(9.8)cos 36 = 17a

208 - 24.6568cos36 = 17a

208 - 19.9478 = 17a

188.05 = 17a

a = 188.05/17

a = 11.06m/s²

<em>Hence the  the magnitude of the resulting acceleration is 11.06m/s²</em>

6 0
2 years ago
A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c
tensa zangetsu [6.8K]

Answer:

E = \frac{3kQ^2}{5R}

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

V = \frac{kq}{r}

now we can say that

q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)

q = \frac{Qr^3}{R^3}

now electric potential is given as

V = \frac{k\frac{Qr^3}{R^3}}{r}

V = \frac{kQr^2}{R^3}

now work done to bring a small charge from infinite to the surface of this sphere is given as

dW = V dq

dW = \frac{kQr^2}{R^3} dq

here we know that

dq = \frac{3Qr^2dr}{R^3}

now the total energy of the sphere is given as

E = \int dW

E = \int_0^R  \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})

E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)

E = \frac{3kQ^2}{5R}

7 0
3 years ago
What term best describes the regular path of a spacecraft or other object around a planetary body?
Law Incorporation [45]
B. Orbit. The planets orbit the sun, the moon orbits earth, etc.
7 0
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