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2 years ago

Answer the following regarding Momentum Change and Force

Physics

1 answer:

Novosadov [1.4K]2 years ago

7 0

The force necessary to gain the velocity is 40 N.

The velocity gained by the object at the end of the second minute is 30 m/s.

<h3>Force of the object</h3>

F = mv/t

where;

m is mass of the object = 200 kg
v is velocity of the object = 60 m/s
t is time of motion = 5 mins = 300 seconds

F = (200 x 60)/(300)

F = 40 N

<h3>Velocity of the object</h3>

F = mv/t

Ft = mv

v = Ft/m

where;

m is mass = 600 kg
F is force = 150 N
t is time = 2 mins = 120 s

v = (150 x 120)/(600)

v = 30 m/s

Thus, the force necessary to gain the velocity is 40 N.

The velocity gained by the object at the end of the second minute is 30 m/s.

Learn more about force here: brainly.com/question/12970081

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3 years ago

If |v|= 11, |w|= 23, |v-w|= 30, find |v+w|

GenaCL600 [577]

Answer:

$|v+w|=20$

Explanation:

We are given that

|v|=11

|w|=23

|v-w|=30

We have to find the value of |v+w|

|a-b|^2=(a+b)\cdot (a+b)=a^2+b^2-2|a||b|cos\theta

Using the formula

$(30)^2=(11)^2+(23)^2-2(11)(23)cos\theta$

$900=121+529-506cos\theta$

$900-121-529=-506cos\theta$

$250=-506cos\theta$

$cos\theta=-\frac{250}{506}$

$|a+b|^2=|a|^2+|b|^2+2a\cdot bcos\theta$

Using the formula

$|v+w|^2=(11)^2+(23)^2+2(11)(23)\times (-\frac{250}{506})$

$|v+w|^2=400$

$|v+w|^2=(20)^2$

$|v+w|=20$

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3 years ago

A mass attached to a spring oscillates in simple harmonic motion with an amplitude of 10 cm. When the mass is 5.0 cm from its eq

timama [110]

When the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.

<h3>Total energy of the mass</h3>

The total energy possessed by the mass under the simple harmonic motion is calculated as follows;

U = ¹/₂kA²

where;

k is the spring constant
A is the amplitude of the oscillation

<h3>Potential energy of the mass at 5 cm from equilibrium point</h3>

P.E = ¹/₂k(Δx)²

<h3>Kinetic energy of mass</h3>

K.E = U - P.E

K.E = ¹/₂kA² - ¹/₂k(Δx)²

<h3>Percentage of its energy that is kinetic</h3>

$K.E (\%) = \frac{U - P.E}{U} \times 100\%\\\\K.E (\%) =\frac{\frac{1}{2}kA^2 - \frac{1}{2}k(\Delta x)^2 }{\frac{1}{2}kA^2} \times 100\%\\\\K.E (\%) = \frac{A^2 - (\Delta x)^2}{A^2} \times 100\%\\\\K.E (\%) = \frac{10^2 - (10-5)^2}{10^2} \times 100\%\\\\K.E (\%) = \frac{10^2 - 5^2}{10^2} \times 100\%\\\\K.E (\%) = 75\%$

Thus, when the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.

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2 years ago

A tension of 35.0N is applied downward on a 2.50kg mass. What is the acceleration of the mass?

Assoli18 [71]

Answer:

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Explanation:

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3 years ago

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