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3 years ago

Calculate the volume of this regular solid.

Physics

1 answer:

fenix001 [56]3 years ago

6 0

Answer:

$2122.64 cm^3$

Explanation:

Some data in the problem are missing.

Missing values:

Radius of the cylinder: 13 cm

Height of the cylinder: 4 cm

The volume of a cylinder is given by

$V=\pi r^2 h$

where

r is the radius of the base of the cylinder

h is the height of the cylinder

In this problem, we have:

r = 13 cm (radius)

h = 4 cm (height)

Therefore, the volume of the cylinder is:

$V=\pi (13)^2(4)=2122.64 cm^3$

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Two equal forces are applied to a door. The first force is applied at the midpoint of the door; the second force is applied at t

TEA [102]

Answer: The second force applied to the doorknob

Explanation:

The formulae for torque is simple the product of the applied force and the perpendicular distance.

The greater the perpendicular force, the greater the torque assuming a constant value of force.

Applying the force at the doorknob gives for a greater distance between the force and the turning point compared to applying the force at the midpoint of the door ( which is at a shorter distance)

6 0

3 years ago

A spacecraft travels at 1.5 X 108 m/s relative to Earth. A process onboard the

kvasek [131]

Answer:

73.6 minutes

Explanation:

relative time = time interval / √(1 - observer velocity² / speed of light²)

we have relative time. we want time interval.

rearrange

time interval = relative time x √(1 - observer velocity² / speed of light²)

convert 85 mins into seconds

85 x 60 = 5100

1.5 x 10⁸ as a number is 150000000

for c = 299 792 458

time interval = 5100 x √(1 - 150 000 000² / 299 792 458²)

for c = 3 x 10⁸

time interval = 5100 x √(1 - 150 000 000² / 300 000 000²)

time interval = 5100 x 0.866

time interval = 4415.71

divide by 60 for back into minutes

time = 73.6 minutes

4 0

2 years ago

The land between two normal faults moves upward to form a

Leya [2.2K]

<span>The land between two normal faults moves upward to form a

Answer:D</span><span>
fault-block mountain.</span>

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4 years ago

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To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has

Roman55 [17]

Answer:

The magnitude of the acceleration is $a = 0.33 m/s^2$

The direction is $- \r k$ i.e the negative direction of the z-axis

Explanation:

From the question we are that

The mass of the particle $m = 1.8*10^{-3} kg$

The charge on the particle is $q = 1.22*10^{-8}C$

The velocity is $\= v = (3.0*10^4 m/s ) j$

The the magnetic field is $\= B = (1.63T)\r i + (0.980T) \r j$

The charge experienced a force which is mathematically represented as

$F = q (\= v * \= B)$

Substituting value

$F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \ ( 1.63 \r i + 0.980 \r j )T)$

$= 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \ X \ \ \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \ \r j))$

$= (-5.966*10^4 N) \r k$

Note :

$i \ \ X \ \ j = k \\\\j \ \ X \ \ k = i\\\\k \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \ X \ \ j = -i\\\\i \ \ X \ \ k = - j\\$

Now force is also mathematically represented as

$F = ma$

Making a the subject

$a = \frac{F}{m}$

Substituting values

$a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}$

$= (-0.33m/s^2)\r k$

$= 0.33m/s^2 * (- \r k)$

6 0

4 years ago

The form of energy in fossil fuels is what kind of energy?

Nonamiya [84]

Solar energy or sunlight...

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3 years ago

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