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2 years ago

10

Physics

1 answer:

Svetlanka [38]2 years ago

4 0

Answer:

D . A mass of 5 kilograms lifted 5 meters in 10 second

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CAN PEOPLE PLEASE HELP, I NEED TO GET TO THE NEXT LEVEL OR WHATEVER ITS CALLED BUT I NEED 5 BRAINLIEST IN 2 DAYS, I ALREADY HAVE

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3 years ago

Read 2 more answers

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

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Figure A-10 shows the back of a typical bottle of engine oil. The label indicates that this oil is SAE 10W-30, with a classifica

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3 years ago

a canon is firing horizontally from the high wall of a castle located in the midst of a flat plain. if i drop a canon ball from

dsp73

D=44.13, Horizontal velocity is unimportant. The time it takes the cannon ball to fall to the ground is the key to determining the height of the cliff. where that cannon is present.

The speed of any projectile travelling along a Horizontal velocity is known as the horizontal velocity. When a particle or object is launched into the air at an angle other than 90 degrees, it moves along the trajectory path and changes the shape of the curve to a parabolic one.

the speed at which velocity changes over time. Due to its magnitude and direction, acceleration is a vector quantity. The first derivative of velocity with respect to time or the second derivative of position with respect to time are further examples. This is called acceleration.

Distance to the ground is d = 1/2gt^2,

where g is the acceleration rate of gravity (9.80665 m/s^2)

and t = 3 secs.

d = .5×9.80665×9 = 44.13 m.

Learn more about horizontal velocity here

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10 months ago

The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t

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