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Oliga [24]
4 years ago
9

Your friend is constructing a balancing display for an art project. She has one rock on the left ( ms=2.25 kgms=2.25 kg ) and th

ree on the right (total mass mp=10.1 kgmp=10.1 kg ). The distance from the fulcrum to the center of the pile of rocks is rp=0.360 m.rp=0.360 m. Answer the two questions below, using three significant digits. Part A: What is the value of the torque ( ????pτp ) produced by the pile of rocks? (Enter a positive value.)
Physics
1 answer:
Firdavs [7]4 years ago
5 0

Answer:

Torque = 35.60 N.m (rounded off to 3 significant figures.

Explanation:

Given details:

The mass of the rock on the left, ms = 2.25 kg

The total mass of the rocks, mp = 10.1 kg

The distance from the fulcrum to the center of the pile of rocks, rp = 0.360 m

(a) The torque produced by the pile of rock, T = F*rp = m*g*rp

Torque = 9.8*0.360*10.1 = 35.6328

Torque = 35.60 N.m (rounded off to 3 significant figures).

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The deflection of alpha particles in rutherford’s gold foil experiments resulted in what change to the atomic model? the additio
skelet666 [1.2K]
The answer is: The addition of a small dense nucleus at the center of the atom.
5 0
3 years ago
Read 2 more answers
A force of 8 N is used to drag a chair 2.5 metres across a room. Calculate the work done to move the chair.
NNADVOKAT [17]

Answer:

Explanation:

GIVEN

Force (F) = 8 N

Distance (d) = 2.5 metres

Work done = ?

WE know we have the formula

work done = F * d

Work done = 8 * 2.5

                   = 20 Joule

Hope it helps :)

8 0
4 years ago
Read 2 more answers
One carpenter builds 1/5 of a house in 3/4 of a day and the other builds 2/5 of a house in the same amount of time. How many hou
AlladinOne [14]

Answer:

4/5 houses

Explanation:

The speed of the first carpenter is:

(1/5 house) / (3/4 day) = 4/15 house/day

The speed of the second carpenter is:

(2/5 house) / (3/4 day) = 8/15 house/day

Together, their combined speed is:

4/15 house/day + 8/15 house/day = 12/15 house/day

= 4/5 house/day

Per day, the two carpenters build 4/5 of a house.

4 0
3 years ago
The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizo
Norma-Jean [14]

Answer:

0.528m

Explanation:

a)58.7 cm = 0.587 m

Let g = 9.8m/s2. When the frog jumps from ground to the highest point its kinetic energy is converted to potential energy:

E_p = E_k

mgh = mv^2/2

where m is the frog mass and h is the vertical distance traveled, v is the frog velocity at take-off

v^2 = 2gh = 2*9.8*0.587 = 11.5

v = \sqrt{11.5} = 3.4 m/s

b) Vertical and horizontal components of the velocity are

v_v = vsin(\alpha) = 3.4sin(58^0) = 2.877 m/s

v_h = vcos(\alpha) = 3.4cos(58^0) = 1.8 m/s

The time it takes for the vertical speed to reach 0 (highest point) under gravitational acceleration g = -9.8m/s2 is

\Delta t = \Delta v / g = \frac{0 - 2.877}{-9.8} = 0.293s

This is also the time it takes to travel horizontally, we can multiply this with the horizontal speed to get the horizontal distance it travels

s_h = v_ht = 1.8*0.293 = 0.528 m

3 0
3 years ago
A 48.5 kg student runs down the sidewalk and jumps with a horizontal speed of 4.25 m/s onto a stationary skateboard. The student
storchak [24]

<u>Answer:</u>

2.39 kg

<u>Explanation:</u>

There is conservation of momentum here in this problem so we will use the following problem:

m_1u_1+m_2u_2=(m_1+m_2)v

where the mass of the student m_1 is 48.5 kg,

the mass of the skateboard m_2 is m_2 kg,

the initial speed of the student u_1 is 4.25 m/s; and

the speed of the student and skateboard  v is 4.05 m/s.

So substituting the given values in the above formula to get:

(48.5*4.25) + (m_2 * 0) = (48.5 + m_2 ) * 4.05

206.125=196.425+4.05m_2

206.125 - 196.425 = 4.05m_2

m_2=\frac{9.7}{4.05}

m_2=2.39

Therefore, the mass of the skateboard is 2.39 kg.

7 0
3 years ago
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