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3 years ago

13

Which is the SI unit of power

1 answer:

Schach [20]3 years ago

6 0

Answer:

Watt

Explanation:

You might be interested in

What does Δx stand for? What SI units are used to measure it?

pav-90 [236]

Answer:

Displacement. Units of measure metre (m)

Explanation:

Delta x Δx stands for displacement, or change in position.

4 0

3 years ago

3. What is the kinetic energy of a 1300 kg car moving at 26.3 m/s?

Alexxx [7]

Answer:K.E=449598.5j

Explanation:

Kinetic energy of a moving car=1/2mv^2

Where m is the mass of the car

And V is the velocity of the car

K.E=1/2 ×1300×26.3^2

K.E=449598.5j

3 0

3 years ago

A hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake.

ArbitrLikvidat [17]

Answer:

L = 499 m

Explanation:

If we assume that the speed of sound is constant, that travels along a straight line, and that the echo is instantaneous, we can find the total distance travelled by the sound, as follows, just applying the definition of average velocity:

$\Delta x = v_{s} * t = 343 m/s* 2.91 s = 998 m$

If we assume that the time needed to reach to the cliff, is the same used for the return travel, the length of the lake will be exactly half of the total distance calculated:

$l_{lake} = \frac{\Delta x}{2} = \frac{998m}{2} = 499 m$

The length of the lake is 499 m.

8 0

3 years ago

Two horses, Thunder and Misty, are accelerating a wagon 1.3 m/s2. The force of friction is 75 N. Thunder is pulling with a force

just olya [345]

Answer:

1327 kg

Explanation:

So the net force exerted on the wagon would be the sum of forces from 2 horses subtracted by friction force

$F = 1000 + 800 - 75 = 1725 N$

This force results in an acceleration of a = 1.3 m/s2. We can use Newton's 2nd law to calculate the mass of the wagon

$F = ma$

$m = F / a = 1725 / 1.3 \approx 1327 kg$

3 0

4 years ago

A particle with charge −5 µC is located on

Nataly [62]

Answer:

36.25 N

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

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

where:

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

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

r is the separation between the two charges

Moreover:

- The force is repulsive if the two charges have same sign

- The force is attractive if the two charges have opposite sign

In this problem, we have 3 charges:

$q_1=-5\mu C = -5\cdot 10^{-6}C$ is the charge located at $x=+10 cm = +0.10 m$

$q_2=+6\mu C=+6\cdot 10^{-6}C$ is the charge located at $x=-8 cm =-0.08 m$

$q_3=+2\mu C=+2\cdot 10^{-6}C$ is the charge located at $x=-2 cm=-0.02 m$

The force between charge 1 and charge 3 is:

$F_{13}=\frac{kq_1 q_3}{(x_1-x_3)^2}=\frac{(9\cdot 10^9)(5\cdot 10^{-6})(2\cdot 10^{-6})}{(0.10-(-0.02))^2}=6.25 N$

And since the two charges have opposite sign, the force is attractive, so the force on charge 3 is to the right (towards charge 1).

The force between charge 2 and charge 3 is:

$F_{23}=\frac{kq_2 q_3}{(x_2-x_3)^2}=\frac{(9\cdot 10^9)(6\cdot 10^{-6})(2\cdot 10^{-6})}{(-0.08-(-0.02))^2}=30.0 N$

And since the two charges have same sign, the force is repulsive, so the force on charge 3 is to the right (away from charge 2).

So the two forces on charge 3 have same direction (to the right), so the net force is the sum of the two forces:

$F=F_{13}+F_{23}=6.25+30.0=36.25 N$

8 0

4 years ago