Ask question

Ask question

All categories

2 years ago

15

A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by

Paul has magnitude 43.0 N and direction 61.0∘ south of west.
How much work does Paul's force do during a displacement of the crate that is 12.0 m in the direction 22.0∘ east of north?

1 answer:

Delicious77 [7]2 years ago

6 0

Answer:

W = -510.98J

Explanation:

Force = 43N, 61° SW

Displacement = 12m, 22° NE

Work done is given as:

W = F*d*cosA

where A = angle between force and displacement.

Angle between force and displacement, A = 61 + 90 + 22 = 172°

W = 43 * 12 * cos172

W = -510.98J

The negative sign shows that the work done is in the opposite direction of the force applied to it.

You might be interested in

A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.

Volgvan

Answer:

A=0.199

Explanation:

We are given that

Mass of spring=m=450 g= $=\frac{450}{1000}=0.45 kg$

Where 1 kg=1000 g

Frequency of oscillation= $\nu=1.2Hz$

Total energy of the oscillation=0.51 J

We have to find the amplitude of oscillations.

Energy of oscillator= $E=\frac{1}{2}m\omega^2A^2$

Where $\omega=2\pi\nu$ =Angular frequency

A=Amplitude

$\pi=\frac{22}{7}$

Using the formula

$0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2$

$A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398$

$A=\sqrt{0.0398}=0.199$

Hence, the amplitude of oscillation=A=0.199

4 0

2 years ago

The value of specific heat for copper is 390 J/kg⋅C∘, for aluminun is 900 J/kg⋅C∘, and for water is 4186 J/kg⋅C∘.

abruzzese [7]

Answer:

The equilibrium temperature is

21.97°c

Explanation:

This problem bothers on the heat capacity of materials

Given data

specific heat capacities

copper is Cc =390 J/kg⋅C∘,

aluminun Ca = 900 J/kg⋅C∘,

water Cw = 4186 J/kg⋅C∘.

Mass of substances

Copper Mc = 235g

Aluminum Ma = 135g

Water Mw = 825g

Temperatures

Copper θc = 255°c

Water and aluminum calorimeter θ1= 16°c

Equilibrium temperature θf =?

Applying the principle of conservation of heat energy, heat loss by copper equal heat gained by aluminum calorimeter and water

McCc(θc-θf) =(MaCa+MwCw)(θf-θ1)

Substituting our data into the expression we have

235*390(255-θf)=

(135*900+825*4186)(θf-16)

91650(255-θf)=(3574950)(θf-16)

23.37*10^6-91650*θf=3.57*10^6θf- +57.2*10^6

Collecting like terms and rearranging

23.37*10^6+57.2*10^6=3.57*10^6θf+91650θf

8.2*10^6=3.66*10^6θf

θf=80.5*10^6/3.6*10^6

θf =21.97°c

5 0

2 years ago

A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences

NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, $q=3\times 10^{-6}\ C$

Mass of charge particle, $m=2\times 10^{-6}\ C$

Speed, $v=5\times 10^{6}\ m/s$

Acceleration, $a=3\times 10^{4}\ m/s^2$

We need to find the minimum magnetic field that would produce such an acceleration. So,

$ma=qvB\ sin\theta$

For minimum magnetic field,

$ma=qvB$

$B=\dfrac{ma}{qv}$

$B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}$

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0

2 years ago

A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a

vlabodo [156]

Answer:

24.3 degrees

Explanation:

A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:

$a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2$

Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.

Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.

So $gsin\alpha = 4.43cos\alpha$

$\frac{sin\alpha}{cos\alpha} = \frac{4.43}{g} = \frac{4.43}{9.81} = 0.451$

$tan\alpha = 0.451$

$\alpha = tan^{-1}0.451 = 0.424 rad = 0.424*180/\pi \approx 24.3^0$

3 0

2 years ago

Scenario A: 120 J of work is done in 6 s. Scenario B: 160 J of work is done in 8 s. Scenario C: 200 J of work is done in 10 s. W

hodyreva [135]

Using the equation P = W/t to solve your problem .

Thus the answer is all of them use the same amount of power. 20 J.

8 0

2 years ago

Read 2 more answers