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3 years ago

HELPPPPPPP MEEEEEEE ASAP PLZZZZZZZZZ

Physics

2 answers:

aleksandr82 [10.1K]3 years ago

7 0

The answer is 24 atoms.

m_a_m_a [10]3 years ago

5 0

Answer:

12 atoms

Explanation:

Please correct me if I am wrong :)

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Explain the importance of having a support network when trying to achieve a healthy lifestyle. Who supports you when it comes to

Vadim26 [7]

Answer:

Who are the people for you then I can help you format the essay

Explanation:

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2 years ago

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The length of second, s pendulum at a place where gravitational acceleration ]g] is 9.8 m/s

amm1812

O.99 m long .simple pendulum time period is 2s for second formula then use formula T=2pi.rt(lenght/gravity)

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3 years ago

A 7.7 kg sphere makes a perfectly inelastic collision with a second sphere initially at rest. The composite system moves with a

klemol [59]

Answer:

15.4 kg.

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m').................... Equation 1

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.

Given: m = 7.7 kg, u' = 0 m/s (at rest)

Let: u = x m/s, and V = 1/3x m/s

Substitute into equation 1

7.7(x)+m'(0) = 1/3x(7.7+m')

7.7x = 1/3x(7.7+m')

7.7 = 1/3(7.7+m')

23.1 = 7.7+m'

m' = 23.1-7.7

m' = 15.4 kg.

Hence the mass of the second sphere = 15.4 kg

7 0

3 years ago

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A locomotive has a mass of 200,000kg. It is moving at 4.5m/s. Find its momentum .

anastassius [24]

Momentum (p) = mass × velocity

P= 200,000×4.5

P= 900,000 .... answer !!

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3 years ago

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Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10

gayaneshka [121]

Answer:

F = 2123.33N

Explanation:

In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.

$\Sigma \tau=0$

Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.

Then, you obtain the following formula:

$-\tau_l+\tau_p+\tau_{cm}=0$ (1)

τl: torque produced by the left support

τp: torque produced by the person

τcm: torque produced by the center of mass of the wooden

The torque is given by:

$\tau=Fd$ (2)

F: force applied

d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

$-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N$

Where d1, d2 and d3 are distance to the right support.

You solve the equation for F and replace the values of the other parameters:

$F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N$

The force applied by the left support is 2123.33 N

8 0

3 years ago