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11111nata11111 [884]

3 years ago

5

A 110 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.220 m/s. How much work must be d

one on the hoop to stop it

1 answer:

Softa [21]3 years ago

6 0

Answer:

the work that must be done to stop the hoop is 2.662 J

Explanation:

Given;

mass of the hoop, m = 110 kg

speed of the center mass, v = 0.22 m/s

The work that must be done to stop the hoop is equal to the change in the kinetic energy of the hoop;

W = ΔK.E

W = ¹/₂mv²

W = ¹/₂ x 110 x 0.22²

W = 2.662 J

Therefore, the work that must be done to stop the hoop is 2.662 J

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Two tuning forks of frequency 480 hz and 484 hz are struck simultaneously. what is the beat frequency resulting from the two sou

umka21 [38]

Ans: Beat frequency = $f_b$ = 4Hz

Explanation:
The beat frequency is equal to the absolute value of the difference in frequency of the two waves. In other words, the number of beats per second is equal to the difference in frequency. It is due to the destructive and constructive interference. <span>According to this interference, sound will be soft or loud.

Hence. the formula is:
</span>Beat frequency = $f_b = |f_2 - f_1|$
<span>
Since,
</span> $f_1 = 480Hz$
$f_2 = 484Hz$

Therefore,
Beat frequency = $f_b = |484 - 480|$

=> Beat frequency = $f_b = 4Hz$
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7 0

3 years ago

What is the mass of an object that has a weight of 110N ?

fgiga [73]

Weight (W) = 110 N
Acceleration due to gravity (g) = 9.8 m/s^2
Let the mass of the object be m.
By using the formula, W = mg, we get,
110 N = 9.8 m/s^2 × m
or, m = 110 N ÷ 9.8 m/s^2
or, m = 11.2 Kg

<u>Answer:</u>

<em><u>The </u></em><em><u>mass </u></em><em><u>of </u></em><em><u>the </u></em><em><u>object </u></em><em><u>is </u></em><em><u>1</u></em><em><u>1</u></em><em><u>.</u></em><em><u>2</u></em><em><u> </u></em><em><u>Kg.</u></em>

Hope you could get an idea from here.

Doubt clarification - use comment section.

3 0

3 years ago

A 35.4 kg girl is riding on the back of a 15.23 kg cart. the cart and the kid are both moving eastward at 4.25 m/s when she step

Grace [21]

Answer:

The final velocity of the cart is $v_c = 7.02 \ m/s$

Explanation:

From the question we are told that

The mass of the girl is $m_g = 35.4 \ kg$

The mass of the cart is $m_c = 15.23 \ kg$

The speed of the cart and kid(girl) is $v = 4.25 \ m/s$

The final velocity of the girl is $v_g = 3.06 \ m/s$

Let assume that velocity eastward is positive and velocity westward is negative (Note that if we assume vise versa it wouldn't affect the answer )

The total momentum of the system before she steps off the back of the cart

is mathematically evaluated as

$p__{T1}} = (m_g + m_c) * v$

substituting values

$p__{T1}} = (35.4 + 15.23) * 4.25$

$p__{T1}} =215.17 \ kg m /s$

The total momentum after she steps off the back of the cart is mathematically evaluated as

$p__{T2}} = (m_g * v_g ) +( m_c * v_c )$

Where $v_c$ is the final velocity of the cart

substituting values

$p__{T2}} = (35.4 * 3.06 ) +( 15.23 * v_c )$

$p__{T2}} = 108. 324 + 15.23 v_c$

Now according to the law of conservation of momentum

$p__{T1}} =p__{T2}}$

So

$215.17 \ kg m /s = 108. 324 + 15.23 v_c$

=> $v_c = 7.02 \ m/s$

Since the value is positive it implies that the cart moved eastward

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UNO [17]

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