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2 years ago

An 6 kg object accelerating from 17 m/s to 10 m/s. What is the change in momentum of the object?

Physics

1 answer:

Oksana_A [137]2 years ago

3 0

Answer:

the change is = 17 × 10 = 170 =28.33

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A coal - fired power station produces 100MJ of electrical energy when it is supplied with 400MJ of energy from its fuel. The eff

Alexus [3.1K]

Answer:

100 divide by 400 times 100 percent

8 0

2 years ago

A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal

Setler [38]

Answer

given,

mass of the ball = 3 kg

swing in vertical circle with radius = 2 m

work done by the gravity = ?

work done by the tension = ?

Work done by the gravity = - m g Δh

Δ h = 2 + 2 = 4 m

Work done by the gravity = $- 3 \times 9.8 \times 4$

= -117.6 J

work done by gravity is equal to -117.6 J

Work done by tension will be equal to zero.

Zero because tension is always perpendicular to velocity

work done by tension is equal to 0 J

7 0

3 years ago

when a circular plate of metal is heated in an oven, its radius increases at .03 cm/min, at what rate is the area increasing whe

Alinara [238K]

Answer:

Rate of change of area will be $9.796cm^2/min$

Explanation:

We have given rate of change of radius $\frac{dr}{dt}=0.03cm/min$

Radius of the circular plate r = 52 cm

Area is given by $A=\pi r^2$

So $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$

Puting the value of r and $\frac{dr}{dt}$

$\frac{dA}{dt}=2\times 3.14\times 52\times 0.03=9.796cm^2/min$

So rate of change of area will be $9.796cm^2/min$

6 0

2 years ago

What is warmer during the night: land or sea/ocean?

kupik [55]

I do believe the answer is land because the ocean/sea gets cold when its night. So the answer is land

5 0

2 years ago

Read 2 more answers

an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le

storchak [24]

On what:

f (is the focal length of the lens) = ?
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

Using the Gaussian equation, to know where the object is situated (distance from the point).

$\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}$
$\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}$
$\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}$
$\frac{1}{f} = \frac{10}{33.18}$
Product of extremes equals product of means:
$10*f = 1*33.18$
$10f = 33.18$
$f = \frac{33.18}{10}$
$\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark$

6 0

3 years ago