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2 years ago

Will mark ask brainlest

Physics

2 answers:

kvv77 [185]2 years ago

5 0

Answer:

woah yes

Explanation:

Lubov Fominskaja [6]2 years ago

5 0

Its astroid one has more mass than astroid 2. I believe.

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Consider dropping a ball from rest. This ball moves from astate of high gravitational potential energy to one of lowgravitationa

Leni [432]

Answer:

From the negative to the positive cable.

Explanation:

The electrons have negative charge, which means that the negative terminal of the battery will suply the electrons, thus they are present in excess on the negative cable and will jump from it to the positive cable. This current direction is called real current.

3 0

3 years ago

If you swing an object on the end of a string around a circle, the string pulls on the object to keep it moving in a circle. Wha

emmainna [20.7K]

Answer:

Explanation:

4 0

2 years ago

Suppose an object is moving in a straight line at 50 miles/hr. According to Newton's first law of motion, the object will ______

Setler [38]

Answer:

Explanation:

Newton's first law of motion states that a body will remain in its state of rest or if its in motion will continue to move in a straight line, unless its acted upon by an external force.The ability of an object to stay at rest or in motion if its in motion is known as inertia.

Hence the correct option is B.

8 0

3 years ago

A small steel roulette ball rolls around the inside of a 30 cm diameter roulette wheel. It is spun at 150 rpm, but is slows to 6

liraira [26]

Solution :

Given

Diameter of the roulette ball = 30 cm

The speed ball spun at the beginning = 150 rpm

The speed of the ball during a period of 5 seconds = 60 rpm

Therefore, change of speed in 5 seconds = 150 - 60

= 90 rpm

Therefore,

90 revolutions in 1 minute

or In 1 minute the ball revolves 90 times

i.e. 1 min = 90 rev

60 sec = 90 rev

1 sec = 90/ 60 rec

5 sec = $\frac{90}{60}\times 5$

= 75 rev

Therefore, the ball made 75 revolutions during the 5 seconds.

7 0

2 years ago

A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calcula

Luda [366]

Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

$V_{ab}$ = i R

$V_{ab}$ = (0.1832) (65)

$V_{ab}$ = 11.91 Volts

(c)

Power dissipated in the resister R is given as

$P_{R}$ = i²R

$P_{R}$ = (0.1832)²(65)

$P_{R}$ = 2.18 Watt

Power dissipated in the internal resistance is given as

$P_{r}$ = i²r

$P_{r}$ = (0.1832)²(0.5)

$P_{r}$ = 0.0168 Watt

5 0

3 years ago