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Over [174]
3 years ago
9

2. Three super strong teenagers pull a heavy crate across the floor. Dion pulls with a force of 18.5 N towards 0°. Shirley pulls

with a force of 16.5 N towards 30°, while Joan pulls with a force of 19.5 N towards 300°
A. What is the resultant force of the heavy crate?
B. What direction (bearing) is the crate moving?

SHOW WORK
WILL MARK BRAINLIEST!!!

Physics
1 answer:
kogti [31]3 years ago
8 0

Answer:

A) The resultant force is 43.4 [N]

B) The movement of the heavy crate is going to the right and in the negative direction on the y-axis

Explanation:

We need to make a sketch of the different forces acting on the heavy crate.

In the attached image we can see the forces and the sum of the vector with their respective angles.

Forces in the X-axis

Fdionx=18.5N\\\\Fshix=16.5*cos(30)=14.29N\\Fjoanx=19.5*cos(60)=9.75N\\\\Forcex= 18.5 + 14.29 + 9.75 = 42.54 N

Forces in the y-axis

FDiony=0[N]\\Fshirley= 16.5*sin(30)=8.25[N]\\Fjoany=19.5*sin(60)=16.88 [N]\\\\Forcesy=0+8.25-16.88= -8.63[N]

Using the Pythagorean theorem

Tforce=\sqrt{(42.54)^{2} +(8.63)^{2} } \\\\Tforce= 43.4N

The movement of the heavy crate is going to the right and in the negative direction on the y-axis, this can be easily seen in the graphical sum of vectors.

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Which of the following is a small star that has reached the end of its stellar evolution?
Akimi4 [234]
D.) White Dwarf

It is the smallest star whose mass is approximately equal or greater than 1.4M
Here, M = mass of the Sun.

Hope this helps!
3 0
3 years ago
Read 2 more answers
If a puddle of water is frozen, do the particles in the ice have kinetic energy? Explain.
TEA [102]
They have some but not very much, the particles in the ice are still vibrating just not as much as in water. the only time a substance would have 0 kinetic energy is when that substance is at 0 degrees kelvin(absolute zero) so far no place in the universe has been recorded at absolute zero though
8 0
4 years ago
A spring with spring constant 33N/m is attached to the ceiling, and a 4.8-cm-diameter, 1.5kg metal cylinder is attached to its l
mylen [45]

Answer:

0.423m

Explanation:

Conversion to metric unit

d = 4.8 cm = 0.048m

Let water density be \who_w = 1000 kg/m^3

Let gravitational acceleration g = 9.8 m/s2

Let x (m) be the length that the spring is stretched in equilibrium, x is also the length of the cylinder that is submerged in water since originally at a non-stretching position, the cylinder barely touches the water surface.

Now that the system is in equilibrium, the spring force and buoyancy force must equal to the gravity force of the cylinder. We have the following force equation:

F_s + F_b = W

Where F_s = kxN is the spring force, F_b = W_w = m_wg = \rho_w V_s g is the buoyancy force, which equals to the weight W_w of the water displaced by the submerged portion of the cylinder, which is the product of water density \rho_w, submerged volume V_s and gravitational constant g. W = mg is the weight of the metal cylinder.

kx + \rho_w V_s g = mg

The submerged volume would be the product of cross-section area and the submerged length x

V_s = Ax = \pi(d/2)^2x

Plug that into our force equation and we have

kx + \rho_w \pi(d/2)^2x g = mg

x(k + \rho_w g \pi d^2/4) = mg

x = \frac{m}{(k/g) + (\rho_w\pi d^2/4)} = \frac{1.5}{(33/9.8) + (100*\pi * 0.048^2/4)} = 0.423 m

6 0
3 years ago
( Just answer 1 & 2) Please help! This is the last thing I have to do. I will mark brainliest asap!!
natima [27]

Answer:

1. Formula = distance/time

D= 112.0 meters, time =4 secs

Speed = d/t =112.0/4

= 28m/s

2.Formula = distance/time

D= 1/4 x 16093.44 m

= 402.35m

Time = 15secs

Speed= d/t

= 402.35/15

=26m/s

7 0
3 years ago
A tall cylinder contains 20 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t
Zanzabum

Answer:

Pressure, P = 3724 Pa

Explanation:

Given that,

Depth of water, h_w=20\ cm =0.2\ m

Depth of oil, h_o=40-20=20\ cm=0.2\ m

The density of water, d_w=1000\ kg/m^3

The densinty of oil, d_o=900\ kg/m^3

We need to find the gauge pressure at the bottom of the cylinder. So, total pressure is equal to :

P=d_wgh_w+d_ogh_o\\\\P=(d_wh_w+d_oh_o)g\\\\P=(1000\times 0.2+900\times 0.2)\times 9.8\\\\P=3724\ Pa

So, the gauge pressure at the bottom of the cylinder is 3724 Pa.

3 0
3 years ago
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