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castortr0y [4]
3 years ago
5

Why does a force perpendicular to an objects velocity change the direction of the velocity but not its magnitude

Physics
1 answer:
inna [77]3 years ago
3 0

Answer:

According to your question although I think an object undergoing uniform circular motion is moving with a constant speed. Nevertheless, it is accelerating due to its change in direction. The direction of the acceleration is inwards,therefore a force perpendicular to an objects velocity change the direction of the velocity but not its magnitude.

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What will happen if you put a usb drive and a magnet too closely together?
kakasveta [241]
<span>If you put a magnet right next to a USB drive, depending on the strength of the magnet and the amount of steel, nickel or cobalt used in the construction of that particular model of USB drive, the drive would either adhere to, or not adhere to, the magnet. This would cause no other significant effects. The storage of data in solid state form (as in USB drives) is not magnetic in nature, so no deletion or any other damage of the stored data would occur.</span>
6 0
3 years ago
Read 2 more answers
The prediction that alcohol slows reaction time is:_________.
ddd [48]

Answer:

b) directional

Explanation:

The prediction that alcohol slows reaction time is Directional . This is because for alcohol to be known as one which slows reaction time means there have been various hypothesis conducted which supports this.

In this case, formulation of a null hypothesis is usually necessary which means that alcohol does not slow reaction time and another alternative hypothesis that suggests that alcohol slows reaction time.

4 0
3 years ago
Help plzzz!!! I only have 10 minutes to turn in
Lostsunrise [7]
  • The mechanic did 5406 Joules of work pushing the car.

That's the energy he put into the car.  When he stops pushing, all the energy he put into the car is now the car's kinetic energy.

  • Kinetic energy = (1/2) (mass) (speed²)

And there we have it

  • The car's mass is 3,600 kg.
  • Its speed is 'v' m/s .
  • (1/2) (mass) (v²) =  5,406 Joules

(1/2) (3600 kg) (v²) = 5406 joules

1800 kg (v²) = 5406 joules

v² = (5406 joules) / (1800 kg)

v² = (5406/1800) (joules/kg)

= = = = = This section is just to work out the units of the answer:

  • v² = (5406/1800) (Newton-meter/kg)
  • v² = (5406/1800) (kg-m²/s²  /  kg)
  • v² = (5406/1800)  (m²/s²)

= = = = =

v = √(5406/1800)  m/s

<em>v = 1.733 m/s</em>

4 0
3 years ago
During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.
lord [1]

Answer:

(a) F_{sm} = 4.327\times 10^{20}\ N

(b) F_{em} = 1.983\times 10^{20}\ N

(c) F_{se} = 3.521\times 10^{20}\ N

Solution:

As per the question:

Mass of Earth, M_{e} = 5.972\times 10^{24}\ kg

Mass of Moon, M_{m} = 7.34\times 10^{22}\ kg

Mass of Sun, M_{s} = 1.989\times 10^{30}\ kg

Distance between the earth and the moon, R_{em} = 3.84\times 10^{8}\ m

Distance between the earth and the sun, R_{es} = 1.5\times 10^{11}\ m

Distance between the sun and the moon, R_{sm} =  1.5\times 10^{11}\ m

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

F_{G} = \frac{Gmm'_{2}}{r^{2}}                             (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}

F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}

F_{sm} = 4.327\times 10^{20}\ N

(b) The force exerted by the Earth on the Moon is given by eqn (1):

F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}

F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}

F_{em} = 1.983\times 10^{20}\ N

(c) The force exerted by the Sun on the Earth is given by eqn (1):

F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}

F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}

F_{se} = 3.521\times 10^{20}\ N

7 0
3 years ago
A bullet with mass m = 5.21 g is moving horizontally with a speed v = 443 m/s when it strikes a block of hardened steel with mas
AlladinOne [14]

Answer:

0.312 m/s

Explanation:

Elastic collisions conserve momentum and kinetic energy

The velocity of the center of mass will not change. It continues at

0.00521(443) / 14.80521 = 0.155893... ≈ 0.156 m/s

To conserve kinetic energy we can think of the center of mass (CoM) as an ideal spring returning to each mass that strikes it an identical speed of collision in the opposite direction.

The CoM sees the target approach at - 0.156 and will see it depart at 0.156 m/s

A ground based observer sees the target depart at the velocity of the CoM plus the relative velocity .

v = 0.156 + 0.156  = 0.312 m/s

6 0
2 years ago
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