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solniwko [45]
3 years ago
11

What is the focal length of a concave mirror that magnifies, by a factor of +3.2, an object that is placed 30 cm from the mirror

?
A. 23cm
B. 32cm
C. 44cm
D. 46cm
Physics
1 answer:
vazorg [7]3 years ago
8 0

Answer:

C. 44 cm

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p = 30 cm is the distance of the object from the mirror

q is the distance of the image from the mirror

We also know that the magnification is

M=-\frac{q}{p}=+3.2

So

q=-3.2p=-3.2(30)=-96 cm

Substituting into the first equation, we find f:

\frac{1}{f}=\frac{1}{30}-\frac{1}{96}=0.0229 cm^{-1} \rightarrow f=43.7 cm \sim 44 cm

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Here a cat is running at constant speed which is given as 10 km/h for  5s

So here the average speed is defined as total distance moved in total time interval

so here it is given by

v_{avg} = \frac{distance}{time}

since

here speed of cat is constant so it will remain the same

And hence the average speed and instantaneous speed at any instant for this duration will remain the same

so here answer would be

<em>average speed = 10 km/h</em>

<em>instantaneous speed = 10 km/h</em>

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3 years ago
Which is the best procedure to make a permanent magnet?
notka56 [123]

placing a magnetically hard material in a strong magnetic field

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3 years ago
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The cheetah is considered the fastest running animal in the world. Cheetahs can accelerate to a speed of 21.7 m/s in 2.50 s and
viktelen [127]

Answer:

1) 64.2 mi/h

2) 3.31 seconds

3) 47.5 m

4) 5.26 seconds

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 0 m/s

v = Final velocity = 21.7 m/s

s = Displacement

a = Acceleration

1) Top speed = 28.7 m/s

1 mile = 1609.344 m

1\ m=\frac{1}{1609.344}\ miles

1 hour = 60×60 seconds

1\ s=\frac{1}{3600}\ hours

28.7\ m/s=\frac{\frac{28.7}{1609.344}}{\frac{1}{3600}}=64.2\ mi/h

Top speed of the cheetah is 64.2 mi/h

Equation of motion

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow t=\frac{21.7-0}{2.5}\\\Rightarrow a=8.68\ m/s^2

Acceleration of the cheetah is 8.68 m/s²

2)

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{28.7-0}{8.68}\\\Rightarrow t=3.31\ s

It takes a cheetah 3.31 seconds to reach its top speed.

3)

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{28.7^2-0^2}{2\times 8.68}\\\Rightarrow s=47.5\ m

It travels 47.5 m in that time

4) When s = 120 m

s=ut+\frac{1}{2}at^2\\\Rightarrow 120=0\times t+\frac{1}{2}\times 8.68\times t^2\\\Rightarrow t=\sqrt{\frac{120\times 2}{8.68}}\\\Rightarrow t=5.26\ s

The time it takes the cheetah to reach a rabbit is 120 m is 5.26 seconds

8 0
3 years ago
A 20-liter container contains 2.0 moles of oxygen at a pressure of 92 kpa. the average kinetic energy of translation of oxygen m
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From ideal gas law, PV=nRT

where P is the pressure, V is the volume of the container, n is number of moles, R is the gas constant and T is the temperature.

Hence, T=\frac{PV}{nR}=\frac{92*10^{3}*2.0*10^{-3}}{2*3.314}

T= 110.65 k

Kinetic Energy = \frac{3}{2}KT=\frac{3}{2}  (1.38*10^{-23})(110.65)

K.E=  2.2*10^{-21}J

<h3>What is a kinetic energy? </h3>

The energy an object has as a result of motion is known as kinetic energy.

A force must be applied to an object in order to accelerate it. We must put in effort in order to apply a force. After the work is finished, energy is transferred to the item, which then moves at a new, constant speed. Kinetic energy is the type of energy that is transferred and is dependent on the mass and speed attained.

Kinetic energy can be converted into other types of energy and transported between objects. A flying squirrel may run into a chipmunk that is standing still, for instance. Some of the squirrel's initial kinetic energy may have been transferred to the chipmunk or changed into another kind of energy after the collision.

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4 0
2 years ago
You have a grindstone (a disk) that is 95.2 kg, has a 0.399 m radius, and is turning at 93 rpm, and you press a steel axe agains
olya-2409 [2.1K]

Answer:

angular acceleration is -0.2063  rad/s²

Explanation:

Given data

mass m = 95.2 kg

radius r = 0.399 m

turning ω = 93 rpm

radial force N  = 19.6 N

kinetic coefficient of friction  μ = 0.2

to find out

angular acceleration

solution

we know frictional force that is = radial force × kinetic coefficient of friction

frictional force = 19.6 × 0.2

frictional force = 3.92 N

and

we know moment of inertia  that is

γ =  I ×α = frictional force × r

so

γ  = 1/2 mr²α

α  = -2f /mr

α  = -2(3.92) /95.2 (0.399)

α  = - 7.84 / 37.9848 = -0.2063

so angular acceleration is -0.2063  rad/s²

3 0
3 years ago
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