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Eduardwww [97]
3 years ago
12

A 50.0-kg child stands at the rim of a merry-go-round of radius 2.25 m, rotating with an angular speed of 3.30 rad/s.. What is t

he child's centripetal acceleration?
Physics
1 answer:
lara [203]3 years ago
6 0

Answer: the child's centripetal acceleration=24.50 m/s²

Explanation:

Given that mass of child= 50 kg

radius of merry go round= 2.25m

angular speed = 3.30 rad/s

 

Centripetal Acceleration  = v²/ r

  But  V= ωr

So Centripetal Acceleration  = v²/ r =  (ωr)²/ r

=(3.30)² x  (2.25)²/ 2.25 = (3.30)² x  2.25

=24.5025m/s²

=24.50 m/s²

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Find the mass if the force is 18 N and the acceleration is 2 m/s2.
Flura [38]

Answer:

<h3>The answer is 9 kg</h3>

Explanation:

To find the mass given the force and acceleration we use the formula

m =  \frac{f}{a}  \\

where

f is the force

a is the acceleration

We have

f = 18 N

a = 2 m/s²

We have

m =  \frac{18}{2}  \\

We have the final answer as

<h3>9 kg</h3>

Hope this helps you

4 0
3 years ago
A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth
andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

F_{D}=\frac{kq^{2}}{DC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

F_{B}=\frac{kq^{2}}{BC^{2}}

F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

F_{A}=\frac{kq^{2}}{AC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N

The direction of FA is along A to C.

The net force along +X axis

F_{x}=F_{A}Cos45-F_{D}

F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N

The net force along +Y axis

F_{y}=F_{B}-F_{A}Sin45

F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N

The resultant force is given by

F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}

F = 4.03\times10^{7}N

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0
3 years ago
Which simple machines make up a wheel barrow
7nadin3 [17]

Answer:

a lever and a wheel and axle (i guess)

Explanation:

don't have any

4 0
3 years ago
PLEASE ANSWER QUICKLY!!! 4. Identify the part or item associated with the internal-combustion engine that the statement is
marshall27 [118]

Answer:

a. Cylinder head

b. Exhaust valve

c. Engine block

d. Stroke

e. Piston

f. Intake valve

g. Cylinder

h. Combustion chamber

i. Crankshaft

j. Spark plug

Explanation:

If you don’t believe me, look up a diagram of an internal combustion engine.

7 0
3 years ago
Read 2 more answers
. What is the relationship between potential energy, kinetic energy, and speed as the skater moves down and up the U-shaped ramp
kykrilka [37]
Top of the U ramp: potential energy is the highest, while kinetic energy is the lowest

Bottom of the U ramp(aka the curve part): potential energy is the lowest and the kinetic energy is the highest

THEREFORE, PE and KE have an INVERSE RELATIONSHIP.
6 0
3 years ago
Read 2 more answers
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