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sleet_krkn [62]
4 years ago
5

A toy gun uses a spring to project a 5.3 - g soft rubber sphere horizontally. The spring constant is 8.0 N/m, the barrel of the

gun is 15 cm long, and a constant frictional force of 0.032 N exists between barrel and projectile. With what speed does the projectile leave the barrel if the spring was compressed 5.0 cm for this launch?
Physics
1 answer:
Vikki [24]4 years ago
7 0

Answer:

v = 1.16 m/s

Explanation:

As per work energy theorem we know that work done by all forces on the sphere is equal to the change in kinetic energy of the sphere

So here we will have

\frac{1}{2}kx^2 - F_f .d = \frac{1}{2}mv^2

here we know that

F_f = 0.032 N

k = 8 N/m

now we have

\frac{1}{2}(8)(0.05^2) - (0.032)(0.15 + 0.05) = \frac{1}{2}(0.0053)v^2

0.01 - 6.4 \times 10^{-3} = 2.65 \times 10^{-3} v^2

v = 1.16 m/s

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State and explain the law of reflection​
Nastasia [14]
The law of reflection states that when a ray of light reflects off a surface, the angle of incidence is equal to the angle of reflection.
7 0
3 years ago
Two large parallel metal plates carry opposite charges. They are separated by 0.20 m and the p.d. between them is 500 V. What is
Bezzdna [24]

Hi there!

We can use the following relationship between the Potential Difference and the Electric field:

V = E d

V = Potential Difference (500V)
E = Electric Field (V/m)
d = separation between plates (0.2 m)

We can rearrange the equation to solve for the electric field:
E = \frac{V}{d}\\\\

Plug in the given values.

E = \frac{500}{0.2} = \boxed{2500 \frac{V}{m}}

3 0
2 years ago
Cars cross a certain point on the highway in accordance with a Poisson process with rate = 3 per minute. If Al runs across the h
Svetradugi [14.3K]

Answer:

The probability is 0.2212

Solution:

As per the question:

Poisson rate, \lambda = 3/min = \frac{3}{60}\ s = 0.05\ s

Now,

Let

X: time of waiting for the next vehicle on the highway

Now,

To find the probability of the next vehicle to arrive after 10 s

The probability distribution function is given by:

f(x) = \lambda e^{- \lambda x}

Now,

P(X < x) = 1 - e^{- \lambda x}

P(X \leq x) = 1 - e^{- 0.05 x}

For X> 0,

P(X > x) = e^{- \lambda x}

P(X < 5) = 1 - e^{- 5\lambda} = 1 - e^{- 0.25} = 1 - 0.7788 = 0.2212

4 0
3 years ago
(a) How many kilograms of water must evaporate from a 60.0-kg woman to lower their body temperature by 0.750ºC?
Mrrafil [7]

Answer:

69.69 g

Explanation:

Evaporation of water will take out latent heat of vaporization.  Let the mass of water be m and latent heat of vaporization of water be 2260000 J per kg

Heat taken up by evaporating water

= 2260000 x m J

Heat lost by body

= mass x specific heat of body x drop in temperature

60 x 3500 x .750  ( specific heat of human body is 3.5 kJ/kg.k)

= 157500 J

Heat loss = heat gain

2260000 m= 157500

m = .06969 kg

= 69.69 g

5 0
4 years ago
Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 20 mph and half the d
Gnoma [55]

Answer:

On the way to grandmother´s, the average speed was 30 mph. On the way back, the average speed was 40 mph.

Explanation:

The average speed is given by the variation of the position over time.

Mathematically:

ΔX / Δt = v

where:

ΔX = distance (final position - initial position)

Δt = time (final time - initial time)

v = speed

On the way to Grandmother´s, we can calculate how much time Julie drove at each speed:

ΔX / Δt = v

ΔX / v = Δt

50 mi / 20 mph = 2.5 h

In the same way, we can calculate how much time she drove at 60 mph:

50 mi / 60 mph = 0.83 h

In total, she drove a distance of 100 mi in (2.5 h + 0.83 h) 3.33 h. Then, the average speed on the way to Grandmother´s was:

<u>ΔX / Δt = v = 100 mi / 3.33 h = 30 mph</u>

In the return trip, we do not know the distance nor the time that she drove at each speed, but we know that for each part of the trip, the time is the same (Δt)  and we also know that the total distance is 100 mi and the total time is 2Δt:

v1 = ΔX1 / Δt

v2 = ΔX2 / Δt

ΔX2 + ΔX1  = 100

where

v1 = speed during the first part of the trip (20 mph)

v2 = speed during the second part of the trip (60 mph)

ΔX1 = distance driven at 20 mph

ΔX2 = distance driven at 60 mph

Δt = time

If we divide v2/v1, we will get:

v2/v1 = (ΔX2 / Δt) / (ΔX1 / Δt)

60 mph / 20 mph = ΔX2 / ΔX1

3 = ΔX2 / ΔX1

3ΔX1 = ΔX2

Then we can replace ΔX2 for 3ΔX1 in this equation:

ΔX2 + ΔX1  = 100 mi

3ΔX1 + ΔX1 = 100 mi

4ΔX1 = 100 mi

ΔX1 = 25 mi

And now, we can solve Δt from the equation of v1:

v1 = ΔX1 / Δt

Δt = ΔX1 / v1 = 25 mi / 20 mph = 1.25 h

The average speed on the return trip is then:

<u>v = 100 mi / 2Δt = 100 mi / 2.5 h = 40mph</u>

8 0
4 years ago
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