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grigory [225]
2 years ago
11

A capacitor is charges with 9.6 nC and has a 120 V potential difference between its terminals. Compute

Physics
1 answer:
IRINA_888 [86]2 years ago
4 0

Explanation:

Q = CV

where C = capacitance

V = potential difference

Solving for C,

C = Q/V = (9.6×10^-9)(120 v)

= 1.15 microFarads

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The cheetah can run a distance of 275 m for 9 seconds. Calculate its speed.
AnnZ [28]

Answer: 110000

Explanation:

26/9=30.5555555556

30.5555555556 x 60=1833.33333333

110000 x 60=110000

4 0
2 years ago
1. A soccer ball is kicked horizontally off a cliff with an initial speed of 8 m/s and lands 16 m from the base of
lana66690 [7]

Answer:

Height of cliff = S = 20 m (Approx)

Explanation:

Given:

Initial velocity = 8 m/s

Distance s = 16 m

Starting acceleration (a) = 0

Computation:

s = ut + 1/2a(t)²

16 = 8t

t = 2 sec

Height of cliff = S

Gravitational acceleration = 10 m/s

S = 1/2a(t)²

S = 1/2(10)(2)²

Height of cliff = S = 20 m (Approx)

3 0
3 years ago
Write a description of this race using these words.
podryga [215]

Answer:

Race refers to physical differences that groups and cultures consider socially significant. For example, people might identify their race as Aboriginal, African American or Black, Asian, European American or White, Native American, Native Hawaiian or Pacific Islander, Māori, or some other race.

Explanation:

hope this helps

8 0
2 years ago
MATHPHYSSSSSSSS PLEASEEEEEE IM SORRY YOU PROBABLY HATE ME
inysia [295]

Answer:

3.1 m/s

Explanation:

First, find the time it takes for the cat to land.  Take down to be positive.

Given:

Δy = 0.61 m

v₀ = 0 m/s

a = 9.81 m/s²

Find: t

Δy = v₀ t + ½ at²

(0.61 m) = (0 m/s) t + ½ (9.81 m/s²) t²

t = 0.353 s

Now find the horizontal velocity needed to travel 1.1 m in that time.

Given:

Δx = 1.1 m

a = 0 m/s²

t = 0.353 s

Find: v₀

Δx = v₀ t + ½ at²

(1.1 m) = v₀ (0.353 s) + ½ (0 m/s²) (0.353 s)²

v₀ = 3.1 m/s

3 0
2 years ago
On the Moon the acceleration due to gravity is about one sixth that on Earth. If a golfer on the Moon imparted the same initial
swat32

Explanation:

We know that that the range of the ball on the earth

R_{earth}=\frac{v_o^2sin2\theta}{g_{earth}}

therefore, range of the ball on moon

R_{moon}=\frac{v_o^2sin2\theta}{g_{moon}}

R_{moon}=\frac{v_o^2sin2\theta}{g_{earth}/12}

therefore,

R_{moon}=6R_{earth}

Therefore, the range of ball will be 6 times on the moon than that on earth

6 0
2 years ago
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