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julsineya [31]
3 years ago
12

What is the voltage across resistor #2? (must include unit - V)

Physics
1 answer:
bixtya [17]3 years ago
7 0

Answer: 120V

Explanation: In a parallel circuit, the voltage is the same and resistance doesn't matter.

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At the top of a looped section of roller coaster track, the car and rider are completely upside down. Engineers calculated that
Pie

Answer:

m v^2 / R = m g where gravitational force provides centripetal force

R = v^2 / g = 14.3^2 m/s / 9.8 m/s^2 = 20.9 m

7 0
3 years ago
9. A radioisotope has a half-life of 4.50 min and an initial decay rate of 8400 Bq. What will be
Akimi4 [234]

Answer:

525 Bq

Explanation:

The decay rate is directly proportional to the amount of radioisotope, so we can use the half-life equation:

A = A₀ (½)^(t / T)

A is the final amount

A₀ is the initial amount,

t is the time,

T is the half life

A = (8400 Bq) (½)^(18.0 min / 4.50 min)

A = (8400 Bq) (½)^4

A = (8400 Bq) (1/16)

A = 525 Bq

8 0
3 years ago
What causes one plate to sink under another <br> 1 Mass<br> 2 Shape<br> 3 Density
solong [7]
It is either the mass or density, I believe it is 3 density tho but it could be mass good luck:)
3 0
2 years ago
Which is an example of the force of attraction between two objects that have mass?
Mariulka [41]

Answer:

Gravity is an example of the force of attraction between two objects that have mass.

4 0
3 years ago
Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates
olasank [31]

2) 20.2 m/s

In the first 4.4 seconds of its motion, the blue car accelerates at a rate of

a=4.6 m/s^2

So its final velocity after these 4.4 seconds is

v=u+at

where

u = 0 is the initial velocity (the car starts from rest)

a is the acceleration

t is the time

Substituting t = 4.4 s, we find

v=0+(4.6)(4.4)=20.2 m/s

After this, the car continues at a constant speed for another 8.5 s, so it will keep this velocity until

t=4.4 + 8.5 = 12.9 s

Therefore, the velocity of the car 10.4 seconds after it starts will still be 20.2 m/s.

3) 216.2 m

The distance travelled by the car during the first 4.4 s of the motion is given by

d_1 = ut_1 + \frac{1}{2}at_1^2

where

u = 0 is the initial velocity

t_1 = 4.4 s is the time

a=4.6 m/s^2 is the acceleration

Substituting,

d_1 = 0 +\frac{1}{2}(4.6)(4.4)^2=44.5 m

The car then continues for another 8.5 s at a constant speed, so the distance travelled in the second part is

d_2 = vt_2

where

v_2 = 20.2 m/s is the new velocity

t_2 = 8.5 s is the time

Substituting,

d_2 = (20.2)(8.5)=171.7 m

So the total distance travelled before the brakes are applied is

d=44.5 m+171.7 m=216.2 m

4) -6.62 m/s^2

We are told that the blue car comes to a spot at a distance of 247 meters from the start. Therefore, the distance travelled by the car while the brakes are applied is

d_3 = 247 m -216.2 m=30.8 m

We can find the acceleration of the car during this part by using the SUVAT equation:

v_f^2 - v_i^2 = 2ad_3

where

v_f = 0 is the final velocity (zero since the car comes to a stop)

v_i = 20.2 m/s is the velocity of the car at the moment the brakes are applied

a is the acceleration

d_3 = 30.8 m

Solving for a, we find

a=\frac{v_f^2 -v_i^2 }{2d}=\frac{0-(20.2)^2}{2(30.8)}=-6.62 m/s^2

5 0
3 years ago
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