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2 years ago

A 60.0 kg object is moving east at 8.00 m/s, and then slows down to 4.00 m/s. How much work was done?

Physics

2 answers:

Artist 52 [7]2 years ago

8 0

The answer is -1440. That is letter A.

almond37 [142]2 years ago

3 0

Answer:

Work done, W = -1440 J

Explanation:

It is given that,

Mas of an object, m = 60 kg

Initial velocity of the object, u = 8 m/s

Final velocity of an object, v = 4 m/s

We know that work done is equal to change in kinetic energy of an object as per work energy theorem.

$W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

$W=\dfrac{1}{2}m(v^2-u^2)$

Putting all the values in above equation we get :

$W=\dfrac{1}{2}\times 60\ kg((4\ m/s)^2-(8\ m/s)^2)$

W = -1440 J

So, the work done by the object is -1440 Joules. Hence, the correct option is (A) " -1440 J ".

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Alexus [3.1K]

Answer:

4,050

Explanation:

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2 years ago

A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t

Pie

Answer:

$I = \frac{mvb}{6}$

Explanation:

we know angular velocity in terms of moment of inertia and angular speed

$L = I$ ω .... (1)

moment of inertia of rod rotating about its center of length b

$I = \frac{ mb^2}{12}$ ........ .(2)

using v = ωr

where w is angular velocity

and r is radius of rod which is equal to b

so we get 2v = ωb

ω = 2v/b ................. (3)

here velocity is two time because two opposite ends are moving opposite with a velocity v so net velocity will be 2v

put second and third equation in ist equation

$L = \frac{mb^2}{12}$ × $\frac{2v}{b}$

so final answer will be $L = \frac{mvb}{6}$

7 0

2 years ago

A bicyclist of mass 68 kg rides in a circle at a speed of 3.9 m/s. If the radius of the circle is 6.5 m, what is the centripetal

ASHA 777 [7]

Data:
Centripetal Force = ? (Newton)
m (mass) = 68 Kg
s (speed) = 3.9 m/s
R (radius) = 6.5 m

Formula:
$F_{centripetal\:force} = \frac{m*s^2}{R}$

Solving:
$F_{centripetal\:force} = \frac{m*s^2}{R}$
$F_{centripetal\:force} = \frac{68*3.9^2}{6.5}$
$F_{centripetal\:force} = \frac{68*15.21}{6.5}$
$F_{centripetal\:force} = \frac{1034.28}{6.5}$
$\boxed{\boxed{F_{centripetal\:force} = 159.12\:N}}$
Answer:
<span>B.159 N</span>

3 0

2 years ago

A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound

NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

$f_n=\frac{nv_s}{4L}$

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

$f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz$

hence, the frequency of the third overtone is 102 Hz

8 0

2 years ago

Tammy leaves the office, drives 26 km due

fredd [130]

Answer:

72.98 km

Explanation:

Her displacement is simply the distance from her final position to her initial position.

Now, I've drawn and attached a triangle diagram to depict this her movement.

Point O is her initial starting point.

Point A is the first point she gets to after travelling north while point B is the final point after travelling north east.

From the triangle, the displacement will be the distance OB which is denoted by x and can be solved from cosine rule.

Thus;

x² = 62² + 26² - 2(62 × 26)cos 120

x² = 4520 + 806

x² = 5326

x = √5326

x = 72.98 km

4 0

2 years ago