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almond37 [142]
3 years ago
10

Ship A is located 4.1 km north and 2.3 km east of ship B. Ship A has a velocity of 22 km/h toward the south and Ship B has a vel

ocity of 40 km/h in a direction 38° north of east. What are the (a) x-component and (b) y-component of the velocity of A relative to B?
Physics
1 answer:
castortr0y [4]3 years ago
6 0

Answer:

a) x component  = -31.25 km/hr

b) y component = 46.64 km/hr

Explanation:

Given data:

A position is 4km north and 2.5 km east to B

Ship A velocity = 22 km/hr

ship B velocity = 40 km/hr

A velocity wrt to velocity of B

\vec{V_{AB}} =\vec{V_A} - \vec{V_B}

\vec{V_A} = 22 km/hr

\vec{V_B} = 40 cos38\hat{i} + 40sin 38 \hat{j}

                 = 31.52\hat{i} + 24.62 \hat{j}

putting respective value to get velocity of  A with respect to B

\vec{V_{AB}} = -22 \hat {j} - (31.52\hat{i} + 24.62 \hat{j})

\vec{V_{AB}} = -31.52\hat{i} - 46.62\hat{j}

a) x component  = -31.25 km/hr

b) y component = 46.64 km/hr

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3 years ago
A 10.0g piece of copper wire, sitting in the sun reaches a temperature of 80.0 C. how many Joules are released when the copper c
Zolol [24]

Answer:

150.8 J

Explanation:

The heat released by the copper wire is given by:

Q=mC_s \Delta T

where:

m = 10.0 g is the mass of the wire

Cs = 0.377 j/(g.C) is the specific heat capacity of copper

\Delta T=40.0 C - 80.0 C=-40.0 C is the change in temperature of the wire

Substituting into the equation, we find

Q=(10.0 g)(0.377 J/gC)(-40.0^{\circ})=-150.8 J

And the sign is negative because the heat is released by the wire.

6 0
3 years ago
Which of the following is the closest to the scientific fact
Finger [1]
B - A theory seems to be the closest
7 0
3 years ago
Mark and David are loading identical cement blocks onto David’s pickup truck. Mark lifts his block straight up from the ground t
Pepsi [2]

Answer:

b) true. The jobs are equal

Explanation:

The work on a body is the scalar product of the force applied by the distance traveled.

    W = F. d

Work is a scalar, the work equation can be developed

    W = F d cos θ

Where θ is the angle between force and displacement

Let's apply these conditions to the exercise

a) False, if we see the expression d cosT is the projection of the displacement in the direction of the force, so there may be several displacement, but its projection is always the same

b) true. The jobs are equal dx = d cosθ

c) False, because the force is equal and the projection of displacement is the same

d) False, knowledge of T is not necessary because the projection of displacement is always the same

e) False mass is not in the definition of work

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3 years ago
A spherical Christmas tree ornament is 8.00 cm in diameter. What is the magnification of an object placed 12.0 cm away from the
LiRa [457]

The magnification of the ornament is 0.25

To calculate the magnification of the ornament, first, we need to find the image distance.

Formula:

  • 1/f = u⁻¹+v⁻¹.................... Equation 1

Where:

  • f = Focal length of the ornament
  • u = image distance
  • v = object distance.

make u the subject of the equation

  • u = fv/(f+v)................ Equation 2

From the question,

Given:

  • f = 8/2 = 4 cm
  • v = 12 cm

Substitute these values into equation 2

  • u = (12×4)/(12+4)
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  • u = 3 cm.

Finally, to get the magnification of the ornament, we use the formula below.

  • M = u/v.................. Equation 3

Where

  • M = magnification of the ornament.

Substitute these values above into equation 3

  • M = 3/12
  • M = 0.25.

Hence, The magnification of the ornament is 0.25

8 0
3 years ago
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