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3 years ago

I need help this is for economics, can anyone help me ?

Physics

1 answer:

PolarNik [594]3 years ago

4 0

i can when do u need it by working on it now!!!

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Dense water near the poles sinks, creating a current towards the equator. What would you expect to happen to this current if tem

Rudik [331]

It's not c or d :/!!!!!

3 0

3 years ago

A CAR ACCElerates FROM REST To

Ilia_Sergeevich [38]

Answer:

666.6 kW

Explanation:

Power Formula

Power = Force × Velocity

Calculating Force

F = ma
F = m(v - u / t) [From the equation v = u + at]
F = 2,000 (50/7.5)
F = 2,000 (20/3)
F = 40,000/3
F = 13333.3 N

Solving for Power

P = 13333.3 × 50
P = 666666.6 W
P = 666.6 kW

5 0

2 years ago

Question 2.

choli [55]

Explanation:

1 inch = 25.4 mm

1 foot = 12 inches

1 mile = 5260 feet

1 cm = 0.01 m or 10 mm

Now Sammy's height is 5 feet and 5.3 inches.

(a) We need to find Sammy's height in inches.

Since, 1 foot = 12 inches

5 feet = 5 × 12 inches = 60 inches

Now, 5 feet and 5.3 inches = 60 inches + 5.3 inches = 65.3 inches

Sammy's height is 65.3 inches.

(b) We need to find Sammy's height in feet.

Since, 1 foot = 12 inches

$1\ \text{inch}=\dfrac{1}{12}\ \text{feet}$

So,

$5.3\ \text{inch}=\dfrac{5.3}{12}\ \text{feet}=0.4416\ \text{feet}$

5 feet and 5.3 inches = 5 feet + 0.4416 feet = 5.44 feet

Sammy's height is 5.44 feet.

3 0

3 years ago

When light passes into a more dense material, it bends away from the

asambeis [7]

Answer: true

Explanation:

7 0

3 years ago

Read 2 more answers

a force 2.4E2 N exists between a positive charge of 8E-5 C and a positive charge of 3E-5 C. What distance separates the charges?

Verdich [7]

The distance between the two charges is $0.3 m$

Explanation:

The electrostatic force between two charged objects is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges of the two objects

r is the separation between the two charges

In this problem, we are given the following:

$q_1 = 8\cdot 10^{-5} C$

$q_2 = 3\cdot 10^{-5} C$

$F=2.4\cdot 10^2 N$

Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

$r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(8.99\cdot 10^9)(8\cdot 10^{-5})(3\cdot 10^{-5})}{2.4\cdot 10^2}}=0.3 m$

Learn more about electrostatic force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0

3 years ago