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noname [10]
2 years ago
10

Richardson pulls a toy 3.0 m across the floor by a string, applying a force of0.50 N. During the first meter, the string is para

llel to the floor. In the next two meters, the string makes an angle 0f 300 with the horizontal direction. What’s the total amount of work done by Richardson on the toy?
Physics
1 answer:
Anastasy [175]2 years ago
7 0

Answer:

Total Work done =0.65 joule

Explanation:

Work done is given Mathematically as

W=F *d

Where w=work done in joules

F=applied force

d= distance moved

The work done to move the toy accros the first meter is

W1=0.5*1

W1=0.5joule

The work done to move the toy across the next 2m at an angle of 30° is

.W2=0.5*2cos30

W2=0.5*2*0.154

W2=0.154joule

Hence total work done is

W1+W2=0.5+0.154

Total Work done =0.65 joule

You might be interested in
a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres
shusha [124]

Answer:

Part a) When collision is perfectly inelastic

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b) When collision is perfectly elastic

v_m = \frac{m + M}{2m}\sqrt{5Rg}

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

mv_m = (m + M)v

so we have

v = \frac{mv_m}{m + M}

now we know that in order to complete the circle we will have

v = \sqrt{5Rg}

\frac{mv_m}{m + M} = \sqrt{5Rg}

now we have

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b)

Now we know that collision is perfectly elastic

so we will have

v = \frac{2mv_m}{m + M}

now we have

\sqrt{5Rg} = \frac{2mv_m}{m + M}

v_m = \frac{m + M}{2m}\sqrt{5Rg}

6 0
2 years ago
You throw a baseball at an angle of 30.0∘∘ above the horizontal. It reaches the highest point of its trajectory 1.05 ss later. A
garri49 [273]

Answer:

The speed with which the baseball leaves the hand = 20.58 m/s

Explanation:

The time take to reach highest height during a projectile's flight is given by

t = (u sin θ)/g

u = initial velocity of the baseball = ?

θ = angle of throw above the horizontal

g = acceleration due to gravity = 9.8 m/s²

1.05 = (u sin 30)/9.8

u = (1.05 × 9.8)/0.5

u = 20.58 m/s

7 0
2 years ago
When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t
antoniya [11.8K]

Answer:

Intensity of beam 18 feet below the surface is about 0.02%

Explanation:

Using Lambert's law

Let dI / dt = kI, where k is a proportionality constant, I is intensity of incident light and t is thickness of the medium

then dI / I = kdt

taking log,

ln(I) = kt + ln C

I = Ce^kt

t=0=>I=I(0)=>C=I(0)

I = I(0)e^kt

t=3 & I=0.25I(0)=>0.25=e^3k

k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

Intensity of beam 18 feet below the surface is about 0.2%

3 0
2 years ago
At one point in the rescue operation, breakdown vehicle A is exerting a force of 4000 N and breakdown vehicle B is exerting a fo
lukranit [14]

Answer:

1.) Magnitude = 5596 N

2.) Direction = 60 degrees

Explanation: You are given that the breakdown vehicle A is exerting a force of 4000 N at angle 45 degree to the vertical and breakdown vehicle B is exerting a force of 2000 N

Let us resolve the two forces into X and Y component

Sum of the forces in the X - component will be 4000 × cos 45 = 2828.43 N

Sum of the forces in the Y - component will be 2000 + ( 4000 × sin 45 )

= 2000 + 2828.43

= 4828.43 N

The resultant force R will be

R = sqrt ( X^2 + Y^2 )

Substitutes the forces at X component and Y component into the formula

R = sqrt ( 2828.43^2 + 4828.43^2 )

R = sqrt ( 31313752.53 )

R = 5595.87 N

The direction will be

Tan Ø = Y/X

Substitute Y and X into the formula

Tan Ø = 4828.43 / 2828.43

Tan Ø = 1.707106

Ø = tan^-1( 1.707106 )

Ø = 59.64 degree

Therefore, approximately, the magnitude and direction of the resultant force on the truck are 5596 N and 60 degree respectively.

8 0
2 years ago
Kimonoski takes a 9-minute shower every day. The shower uses about 1.8 gal per minute of water. He also uses 23 gallons of hot w
ioda

Answer:

Q_{week} = 458884.6\, BTU

Explanation:

The weekly water consumption of Kimonoski is:

m_{bath,week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (1.8\,\frac{gal}{min} )\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (\frac{1\,min}{60\,s} )\cdot (9\,min)\cdot (\frac{60\,s}{1\,min} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)

m_{bath.week} = 948.205\,lbm

m_{others, week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (23\,gal)\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)

m_{others, week} = 1346.218\,lbm

m_{week} = m_{bath,week} + m_{others, week}

m_{week} = 2294.423\,lbm

The total energy required per week for hot water is:

Q_{week} = m_{week}\cdot c_{p,water}\cdot \Delta T

Q_{week} =(2294.423\,lbm)\cdot (1\,\frac{BTU}{lbm\cdot ^{\textdegree}F} )\cdot (50^{\textdegree}F)

Q_{week} = 458884.6\, BTU

3 0
1 year ago
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