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3 years ago

Which is the SI unit of power

Physics

1 answer:

Schach [20]3 years ago

6 0

Answer:

Watt

Explanation:

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Mashcka [7]

Answer:

Is it 11.8m high or 0.6?

The question is not clear

7 0

3 years ago

The velocity of a ball as it is thrown up is 20 meters/second. It attains zero velocity after 3.5 seconds. What is the average a

Step2247 [10]

Change in speed = (speed at the end) minus (speed at the beginning)

= (0 - 20 m/s) = -20 m/s

Acceleration = (change in speed) / (time for the change)

= (-20 m/s) / (3.5 sec) = -5.71 m/s²

Notice that this story probably didn't happen on Earth, because
the acceleration due to gravity on Earth is 9.8 m/s².

3 0

3 years ago

A cell phone weighing 80 grams is flying through the air at 15 m/s what is the kinetic energy

hjlf

Answer:

$9\:\mathrm{J}$

Explanation:

Kinetic energy is given by the following equation:

$KE=\frac{1}{2}mv^2$ , where $m$ is mass in $\mathrm{kg}$ and $v$ is velocity in $\mathrm{m/s}$ .

Since the cell phone's mass is given in grams, we need to convert this into kilograms:

$80\:\mathrm{g}\cdot \frac{1\:\mathrm{kg}}{1000\:\mathrm{g}}=0.08\:\mathrm{kg}$ .

Therefore, the kinetic energy of the cell phone is:

$KE=\frac{1}{2}\cdot 0.08\cdot 15^2=\fbox{$9\:\mathrm{J}$}$ .

3 0

3 years ago

A wheel of radius 0.23 m, which is moving initially at 25.0 m/s, rolls to a stop in 246.0 m. The wheel's rotational inertia is 0

Digiron [165]

Answer: -1.27 m/s^2

Explanation:

a = - V^2 / 2x

a = -(25^2) / 2 x (246) = 1.27 m/ s^2

Therefore the linear acceleration of the wheel is - 1.27 m/s^2

8 0

3 years ago

An insect 1.1 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens

Pavlova-9 [17]

Answer:

Explanation:

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

lens formula

$\frac{1}{v} -\frac{1}{u} =\frac{1}{f}$

Putting the values

$\frac{1}{v} +\frac{1}{15} =\frac{1}{14}$

v = 210 mm .

B )

magnification = v / u

= 210 / 15

= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

$\frac{210}{15} \times \frac{D}{f_e}$

D = 25 cm , f_e = focal length of eye piece

= 14 x 250 / 21

= 166.67

= 170 ( in two significant figures )

7 0

3 years ago