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olganol [36]
3 years ago
10

Assume it takes 10 J to stretch a spring 10 cm beyond its natural length. Find the work required (in Joules) to stretch the spri

ng a further 10 cm, i.e. find the work required to stretch the spring from 10 cm beyond its natural length to 20 cm beyond its natural length. (Do not enter units)
Physics
1 answer:
Lady bird [3.3K]3 years ago
4 0

Answer:

W = 30 J

Explanation:

given,

Work done = 10 J

Stretch of spring, x = 0.1 m

We know,

dW = F .dx

we know, F = k x

\int dW = \int_0^{0.1} k.x dx

W = \int_0^{0.1} k.x dx

W = k[\dfrac{x^2}{2}]_0^{0.1}

10 = k\dfrac{0.1^2}{2}

k = 2000

now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.

W = \int_{0.1}^{0.2} 2000 x dx

W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx

W = 1000 x 0.03

W = 30 J

Hence, work done is equal to 30 J.

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A car is moving north on a freeway. If a bug is flying south on the freeway, is the momentum of the bug positive or negative?
faltersainse [42]

Negative

Because the car is moving up and the bug is moving down. but it also depends on the weather so choice between one of those two I think is Negative but I may be wrong.

3 0
2 years ago
I drop an egg from a certain distance and it takes 3.74 seconds to reach the ground. How high up was the egg?
OverLord2011 [107]

Answer:

68.5 meters

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 3.74 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0) (3.74) + ½ (9.8) (3.74)²

Δy = 68.5

The egg fell 68.5 meters.

7 0
3 years ago
an ice sheet 5m thick covers a lake that is 20m deep. at what is the temperature of the water at the bottom of the lake?
muminat

Answer:

4°C

Explanation:

Water is densest at 4°C.  Since dense water sinks, the bottom of the lake will be 4°C.

6 0
3 years ago
A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of
balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

f=\mu N

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, N=mg

∴ f=\mu mg=0.490\times 90\times 9.8=432.18\ N

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

W=-fd=-432.18d

Now, change in kinetic energy is given as:

\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J

Therefore, from work-energy theorem,

W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m

Hence, the skater covers a distance of 15 m before stopping.

7 0
3 years ago
A tennis ball is served horizontally from 2.4m above the ground at net is 12m away and point 0.9 high will be ball clear the net
Schach [20]

Explanation:

Let us first calculate  long does it take to go 12m at 30m/s( assumed speed)

12/30 = 0.4 seconds

horizontal distance the ball drop in that time

H= (0)(0.4)+1/2(-9.8)(0.4)2

H= -0.78m

negative sign shows that the height of the ball at the net from the top.

Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m

As 1.62m>0.9m so the ball will clear the net.

H_1= V0y t’ + ½ g t’^2

-2.4= (0)t’ + ½ (-9.8) t’^2

t’= 0.69s

X’=V0x t’

X’=(30)(0.96)

X’= 20.7m

3 0
3 years ago
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