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3 years ago

A block pushed with a force of 27 N accelerates at 5 m/s2. What is the mass of the block?

Physics

1 answer:

Semmy [17]3 years ago

5 0

Explanation:

F=ma

26N=m(5m/s)

=5.2kg

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V1 * sin(θ) where θ is the angle v1 makes with the vertical.

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4 years ago

Ik you didnt do the lab but like idek...

serious [3.7K]

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Speed and direction affect pitch.

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3 years ago

A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg. The free end of the stri

Dmitrij [34]

Answer:

(a). The angular speed of the rotating is 33.8 rad/s.

(b). The speed of its center is 2.7 m/s.

Explanation:

Given that,

Radius = 8.00 cm

Mass = 0.180 kg

Height = 75.0 m

We need to calculate the angular speed of the rotating

Using conservation of energy

$\dfrac{1}{2}I\omega_{1}^2+\dfrac{1}{2}mv_{1}^{2}+mgh_{1}=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}+mgh_{2}$

Here, initial velocity and angular velocity are equal to zero.

$mgh_{1}=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}+mgh_{2}$

$mg(h_{1}-h_{2})=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}$

$mgH=\dfrac{1}{2}mr^2\omega_{2}^2+\dfrac{1}{2}m(r\omega_{2})^2$

Here, $H = h_{1}-h_{2}$

$gH=r^2\omega_{2}^2$

$\omega_{2}^2=\dfrac{gH}{r^2}$

$\omega_{2}=\sqrt{\dfrac{9.8\times75.0\times10^{-2}}{(8.00\times10^{-2})^2}}$

$\omega_{2}=33.8\ rad/s$

The angular speed of the rotating is 33.8 rad/s.

(b). We need to calculate the speed of its center

Using formula of speed

$v=r\omega$

Put the value into the formula

$v=8.00\times10^{-2}\times33.8$

$v=2.7\ m/s$

Hence, (a). The angular speed of the rotating is 33.8 rad/s.

(b). The speed of its center is 2.7 m/s.

3 0

3 years ago

Read 2 more answers

Please help!!!!!!!!!!

wlad13 [49]

Answer:

C. Acceleration due to gravity

Explanation:

which is 9.8 m/s2 on Earth. The formula for calculating weight is F = m × 9.8 m/s2.

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3 years ago

Important minerals such as copper are found in what?

adoni [48]

It can be found in granitic and basaltic rock.

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4 years ago