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2 years ago

6

The desperate contestants on a TV survival show are very hungry. The only food they can see is some fruit hanging on a branch hi

gh in a tree. Fortunately, they have a spring they can use to launch a rock. The spring constant is 1000 N/m, and they can compress the spring a maximum of 30 cm. All the rocks on the island seem to have a mass of 400 g.a. With what speed does the rock leave the spring?b. If the fruit hangs 15 m above the ground, will they feast or go hungry?

1 answer:

emmasim [6.3K]2 years ago

6 0

Answer:

(a) v = 15m/a

(b) No they won't feast because the rock can only rise to a height of 11.5m which is less than 15m.

Explanation:

Please see the attachment below for film solution.

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The muzzle velocity of a 50.0g shell leaving a 3.00 kg rifle is 400m/s what is the recoil velocity of the rifle

serg [7]

Here if we consider bullet and gun as a system then we can say that momentum of the system will remain conserved

so we will have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

now we know that

$m_1 = 50 g = 0.050 kg$

$m_2 = 3 kg$

$v_{1i} = v_{2i} = 0$

$v_{1f} = 400 m/s$

now we will have

$0.050(0) + 3(0) = 0.050(400) + 3(v)$

$20 + 3v = 0$

$v = - \frac{20}{3} = - 6.67 m/s$

so gun will recoil with speed 6.67 m/s

6 0

2 years ago

A photon ionizes a hydrogen atom from the ground state. The liberated electron 11. recombines with a proton into the first excit

anygoal [31]

Answer:

a) 23.2 e V

b) energy of the original photon is 36.8 eV

Explanation:

given,

energy at ground level = -13.6 e V

energy at first exited state = - 3.4 e V

A photon of energy ionized from ground state and electron of energy K is released.

h ν₁ - 13.6 = K

K combine with photon in first exited state giving out photon of energy

$h\nu_2 =\dfrac{hc}{\lambda}=\dfrac{12400}{466}$

= 26.6 e V

h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°

K + ( 3.4 ) = 26.6 e V

a) energy of free electron

K = 26.6 - 3.4 = 23.2 e V

b) energy of the original photon

h ν₁ - 13.6 = K

h ν₁ = 23.2 + 13.6

= 36.8 e V

energy of the original photon is 36.8 eV

3 0

2 years ago

Wind blows at the speed of 30m/s across a 175m^2 flat roof if a house.

Makovka662 [10]

Answer:

the net force is 101587.5 N

Explanation:

The speed of wind

v = 30 m/s

The area of roof,

A = 175 m 2

The expression for the Bernoulli's theorem.

P = 12 ρv 2 ...... (1)

Here,

P is the pressure difference,

ρ is the density of air and

v is the speed of wind.

The expression for the pressure.

P = F A ..... (2)

Here,

F is the force and

A is the area of roof.

Part (a)

Substitute the values for the pressure difference in equation (1)

P = 12 × 1.29 × (30) 2 P = 580.5 Pa

Thus, the pressure difference at the roof between the inside and outside air is

580.5 Pa

Part (b)

Substitute the values for the net force in equation (2)

580.5 = F 175 F = 101587.5 N

Thus, the net force is 101587.5 N.

8 0

2 years ago

A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

$\displaystyle P=\frac{W}{t}$

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0

2 years ago

Collect information about working of Geyser and prepare a report

ikadub [295]

A geyser is actually a devise that coverts electrical energy into heat energy for heating up water. The heating element that is inside the geyser actually gets heated up and then in turn it heats the water in contact with it within the geyser. There is also a thermostat device within the geyser that cuts off the heating when the water temperature reaches the desired level. This helps in stopping of electrical energy loss. One inlet brings in cold water while another outlet gets rid of the hot water. When the temperature of the water falls below the desired level the heating is again started by the thermostat.

6 0

2 years ago