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3 years ago

A 10 kg ball has a momentum of 40 kg.m/s.what is the ball's speed?

Physics

1 answer:

iren [92.7K]3 years ago

7 0

M = 10kg
P = 40 kg.m/s
V= 4m/s

Formula v = p/m

V= p/m
=4okg.m/s / 10kg
=4m/s

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Susan works 8 hours a day and makes $7.00 per hour. How much money does Susan earn in one week if she

photoshop1234 [79]

Get the amount of money in a day, (8)(7)=$56 per day
5 days per week, 56(5)=280
To conclude, she gets $280 in a week.

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3 years ago

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. While a person lifts a book of mass 2 kg from the floor to a tabletop, 1.5 m above the floor, how much work does the the perso

viktelen [127]

Answer:

a) 29.4 J

b) - 29.4 J

Explanation:

Given:

Mass of the book, m = 2 kg

Height above the floor, h = 1.5 m

Now,

the work done by the person will be = Force applied on the book × displacement of the book

thus,

Work done by the person = mg × h

where, g is the acceleration due to gravity

thus, on substituting the values, we get

Work done by the person = 2 × 9.8 × 1.5 = 29.4 J

now,

for the force applied by the gravitational pull (downwards) the displacement is in opposite direction (upwards) to the force of the gravity.

Thus,

work done by the gravity will be negative

therefore, the work done by the gravity = - mg × h

work done by the gravity = - 29.4 J

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2 years ago

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A cardinal (Richmondena cardinalis) of mass 3.70×10−2 kg and a baseball of mass 0.144 kg have the same kinetic energy. What is t

Radda [10]

Answer:

$\frac{p_{c}}{p_{b}}\approx 0.507$

Explanation:

Since the cardinal and ball have the same kinetic energy, it is possible to determine the ratio between speeds. (c for cardinal, b for baseball)

$K_{c} = K_{b}$

$\frac{1}{2}\cdot m_{c}\cdot v_{c}^{2}= \frac{1}{2}\cdot m_{b}\cdot v_{b}^{2}$

$\frac{v_{c}}{v_{b}}=\sqrt{\frac{m_{b}}{m_{c}} }$

The ratio is obtained by multiplying each side by $\frac{m_{c}}{m_{b}}$ :

$\frac{p_{c}}{p_{b}}=\frac{m_{c}}{m_{b}}\cdot \sqrt{\frac{m_{b}}{m_{c}} }$

$\frac{p_{c}}{p_{b}}= \sqrt{\frac{m_{c}}{m_{b}} }$

The value of this ratio is:

$\frac{p_{c}}{p_{b}}\approx 0.507$

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3 years ago

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A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continu

malfutka [58]

Answer:

Acceleration, $a=-2.48\ m/s^2$

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

$v^2-u^2=2as$

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}$

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3 years ago

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Mercury has a high boiling point of 357 degrees C.
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3 years ago