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Ksivusya [100]
3 years ago
12

A load of bricks is lifted to the second floor of a building. How do work and power relate to this job?

Physics
2 answers:
kow [346]3 years ago
6 0

The faster the job is done, the greater the power. Just took the quiz on e2020 got 100%

Nady [450]3 years ago
5 0
The faster the job is done, the greater the power
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A bird flies 2.8 due west and then 2.7 due north. Another bird flies 2.7 due west and 2.8 due north. What is the angle between t
Elenna [48]
Assume a 2D plane and both the birds start at origin. +X is east, -X is west, +Y is North, -Y is South. 
So, after travel position of bird1 is (-2.8, 2.7) and bird2 is (-2.7,2.8);
slope of bird1 is 2.8/2.7
slope of bird2 is 2.7/2/8
Now we have two solpes b1 and b2
formula = (b2-b1)/(1+b1*b2);

7 0
3 years ago
Jennifer is making a collage using her favorite photos. She wants to cover the collage with something that allows her to see the
Karolina [17]
B.) A transparent material
4 0
3 years ago
Read 2 more answers
How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J

In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

5 0
2 years ago
The distance between two charged objects is doubled. What happens to the electrostatic force between the two?a)It will double.b)
zzz [600]

Answer:

d) It will be cut to a fourth of the original force.

Explanation:

The magnitude of the electrostatic force between the charged objects is

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is the separation between the two objects

In this problem, the initial distance is doubled, so

r' = 2r

Therefore, the new electrostatic force will be

F=k\frac{q_1 q_2}{(r')^2}=k\frac{q_1 q_2}{(2r)^2}=\frac{1}{4}(k\frac{q_1 q_2}{r^2})=\frac{1}{4}F

So, the force will be cut to 1/4 of the original value.

5 0
3 years ago
A car traveling with constant speed travels 150 km in 7200 s what is speed of the car
spayn [35]

The speed of the car is exactly 150/7200 km/sec, or 125/6 meters/sec. 

In more familiar units, that speed is equivalent to ...

-- (20 and 5/6) meters/sec

-- 75 km/hour

8 0
3 years ago
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