Which layer of the sun can be seen during a solar eclipse and contains jets of rapidly moving gas called spicules?

A light spring of constant 163 N/m rests vertically on the bottom of a large beaker of water.

LUCKY_DIMON [66]

Answer:

24.71cm

Explanation:

We approach this problem base don Hooke's law which states the elongation produced in an elastic material is proportional to the applied load or force provided that its elastic limit is not exceeded. This is expressed mathematically as follows;

$F=ke................(1)$

where F is the applied force, k is the force constant and e is the elongation or extension of the material.

In this problem, the applied force F is the weight of the wood which is calculated as follows,

$F=mg.............(2)$

m = 4.11kg

$g=9.8m/s^2$

Hence,

$F=4.11*9.8\\F=40.278N$

Given that k = 163N/m, we make appropriate substitutions into equation (1) to obtain the following;

$40.278=163*e\\e=\frac{40.78}{163}\\e=0.2471m$

Since it is required in cm, we perform the conversion as follows, knowing that 100cm = 1m

$0.2471m=0.2471*100cm=24.71cm$

NB: We do not necessarily need the the density of the wood to perform our calculations since other parameters were given from which we were able to obtain its weight.

7 0

4 years ago

A 1000 kg red car traveling at 40 m/s rear ends a 3000 kg blue car traveling at 35 m/s. the cars bounce off each other and the b

klemol [59]

Answer:

32.5 ms⁻¹

Explanation:

Theory

The law of conservation of linear momentum

The sum of linear momentum of a system under no external force remains constant.

In this scenario although different forces act on two vehicles by each other when we consider the total system of two cars no external unbalance force act on them. So we can apply this law.

As law says,

The sum of linear momentum before collision = The sum of linear momentum after collision.

Momentum = mass × velocity

1000×40+3000×35 = 1000×v + 3000×37.5

where v = velocity of the red car after the cash

v = 32.5 ms⁻¹

8 0

3 years ago

A ball, initially at rest, reaches a speed of 20 m/s at the bottom of the apparatus. If it takes the ball 10 seconds to reach th

MariettaO [177]

Answer:

2 m/s²

Explanation:

Given:

v₀ = 0 m/s

v = 20 m/s

t = 10 s

Find: a

a = (v − v₀) / t

a = (20 m/s − 0 m/s) / 10 s

a = 2 m/s²

3 0

3 years ago

A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to th

kari74 [83]

Given that,

Mass = 9.2 kg

Force = 110 N

Angle = 20.2°

Distance = 5.10 m

Speed = 1.58 m/s

(A). We need to calculate the work done by the gravitational force

Using formula of work done

$W_{g}=mgd\sin\theta$

Where, w = work

m = mass

g = acceleration due to gravity

d = distance

Put the value into the formula

$W_{g}=9.2\times(-9.8)\times5.10\sin20.2$

$W_{g}=-158.8\ J$

(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction

Using formula of potential energy

$\Delta U=-W$

Put the value into the formula

$\Delta U=-(-158.8)\ J$

$\Delta U=158.8\ J$

(C). We need to calculate the work done by 100 N force on the crate

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=100\times5.10$

$W=510\ J$

We need to calculate the work done by frictional force

Using formula of work done

$W=-f\times d$

$W=-\mu mg\cos\theta\times d$

Put the value into the formula

$W=-0.4\times9.2\times9.8\cos20.2\times5.10$

$W=-172.5\ J$

We need to calculate the change in kinetic energy of the crate

Using formula for change in kinetic energy

$\Delta k=W_{g}+W_{f}+W_{F}$

Put the value into the formula

$\Delta k=-158.8-172.5+510$

$\Delta k=178.7\ J$

(E). We need to calculate the speed of the crate after being pulled 5.00m

Using formula of change in kinetic energy

$\Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})$

$v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2$

Put the value into the formula

$v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58$

$v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}$

$v_{2}=6.35\ m/s$

Hence, (A). The work done by the gravitational force is -158.8 J.

(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.

(C). The work done by 100 N force on the crate is 510 J.

(D). The change in kinetic energy of the crate is 178.7 J.

(E). The speed of the crate after being pulled 5.00m is 6.35 m/s

8 0

3 years ago

A ball is fired at an angle of 45 degrees, the angle that yields the maximum range in the absence of air resistance. What is the

Nesterboy [21]

Look at the picture for the answer

7 0

4 years ago

Read 2 more answers