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2 years ago

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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.

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The same physics student jumps off the back of her Laser again, but this time the Laser is

soldi70 [24.7K]

a) The speed of the student after the jump is 1.07 m/s

b) The final speed of the laser is 10.4 m/s

Explanation:

a)

We can solve this problem by applying the law of conservation of momentum: if there are no external forces acting on the system, the total momentum of the student+Laser system must be constant. Therefore, we can write:

$p_i = p_f\\0=mv+MV$

where

The initial momentum is zero

m = 42 kg is the mass of the Laser

v = 1.5 m/s is the final velocity of the Laser

M = 59 kg is the mass of the student

V is the final velocity of the student

Solving the equation for V, we find the velocity of the student:

$V=-\frac{mv}{M}=-\frac{(42)(1.5)}{59}=-1.07 m/s$

So, the final speed of the student is 1.07 m/s.

b)

In this case, the laser and the student are travelling at 3.1 m/s before the student jumps off: therefore, the total momentum before the jump is not zero.

So, the equation of the conservation of momentum is

$(m+M)u=mv+MV$

where

m = 42 kg is the mass of the Laser

M = 59 kg is the student's mass

u = 3.1 m/s is the initial velocity of the student and the Laser

V = -2.1 m/s is the velocity of the student after the jump (she jumps backward)

v is the final velocity of the Laser

And solving for v, we find

$v=\frac{(m+M)u-MV}{m}=\frac{(42+59)(3.1)-(59)(-2.1)}{42}=10.4 m/s$

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

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#LearnwithBrainly

3 0

2 years ago

Read 2 more answers

An electron moves through a uniform electric field vector E = (2.80î + 5.20ĵ) V/m and a uniform magnetic field vector B = 0.400k

alina1380 [7]

Answer:

1.758820×10^11(-2.5i-0.8j) m/s^2

Explanation:

From the question, the parameters given are; E=(2.80i+ 5.20j) v/m, a uniform magnetic field,B= 0.400K T, acceleration, a= ??? and velocity vector, v= 11.0i metre per seconds (m/s)...

We can solve this problem using the formula below;

Ma= q[E+V × B] ---------------(1).

Note: q is negative, m= mass of electron.

Making acceleration,a the subject of the formula and substituting the parameters into equation (1);

a= -e/m × (2.5i + 5.2j +11.0i × 0.400K)

a= -e/m × (2.5i+5.2j-4.4j)

a= e/m × (-2.5i - 0.8j)

e/m= 1.758820×10^11 c/kg

Therefore, slotting in the value of charge to mass(e/m) ratio;

a= 1.7588×10^11×(-2.5i-0.8j) m/s^2

7 0

2 years ago

A truck is driving over a scale at a weight station. When the front wheels drive over the scale, the scale reads 5800 N. When th

aev [14]

Answer:

$x_2=1.60m$

Explanation:

From the Question We are told that

Initial Force $F_1=5800N$

Final Force $F_2=6500N$

Distance between the front and rear wheels \triangle x=3.20 m

Since

$\triangle x=3.20 m$

Therefore

$x_1+x_2=3.20$

$x_1=3.20-x_2$

Generally the equation for The center of mass is at x_2 is mathematically

given by

$x_2 =\frac{(F_1x_1+F_2x_2)}{(F_1+F_2)}$

$x_2=3.20F_1-\frac{x_2F_1+F_2x_2}{(F_1+F_2)}$

$2*F_1*x_2 =3.20F_1$

$x_2=1.60m$

6 0

1 year ago

Two atoms collide while moving in a dilute gas. The larger atom has a mass M1 = 6 Daltons and a speed v1 = 200 m/s, while the sm

olga2289 [7]

Answer:

a). ΔP1=-2.4 $x10^{3} \frac{D*m}{s}$

b). Pp=0 F=0

c). ΔP2=2.4 $x10^{3} \frac{D*m}{s}$

Explanation:

Initial momentum

$P_{1}=m_{1}*v_{i1}$

Final momentum

$P_{1f}=m_{1}*v_{f1}=-m_{1}*v_{i1}$

The change of momentum m1 is:

a).

ΔP1= $P_{1f}-P_{1}$

ΔP1= $-m_{1}*v_{i1}-m_{1}*v_{i1}$

ΔP1= $-2*m_{1}*v_{i1}$

ΔP1= $-2*6 D*200\frac{m}{s}$

ΔP1= $-2.4x10^{3}\frac{D*m}{s}$

b).

The law of conservation of energy in this case there is not external forces so the momentum of the pair change is equal to zero

P=0

Fx=0

c).

ΔP1+ΔP2=0

ΔP2=-ΔP1

ΔP2=- $-2.4x10^{3}\frac{D*m}{s}$

ΔP2= $2.4x10^{3}\frac{D*m}{s}$

8 0

2 years ago

Determine the minimum number of photons that must enter the eye's pupil each second in order for an object to be seen. Assume th

irina1246 [14]

Answer:

Explanation:

Energy of a photon = hν , h is plank's constant and ν is frequency .

= h c / λ ; c is velocity of light and λ is wavelength .

= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 550 x 10⁻⁹

= .036 x 10⁻¹⁷ J

Let no of photons be n .

Total energy of n photons

= n x .036 x 10⁻¹⁷ J

Intensity of light = 5 x 10⁻¹² J /m²

Total energy flowing through eye per second

= intensity x area of eye.

= 5 x 10⁻¹² x π x .2² x 10⁻⁴

= .628 x 10⁻¹⁶ J

so

n x .036 x 10⁻¹⁷ = .628 x 10⁻¹⁶

n = .628 x 10⁻¹⁶ / .036 x 10⁻¹⁷

= 17.36 x 10

= 173.6

174.

8 0

2 years ago