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fgiga [73]
3 years ago
11

The toy car is about 3 inches long is an example of a ?

Physics
1 answer:
balandron [24]3 years ago
8 0

Answer:

a quantitative observation because it includes numerical data

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Assignment
Igoryamba

Answer:

step bro was stuck on the elevator

Explanation:

5 0
3 years ago
Give me four reasons pluto is a cool planet / dwarf planet
AleksandrR [38]

Answer:

1. Just because it is small doesn't mean it needs to be excluded. If that were the case I would've been out of my friend group a while ago

2. Look at it it's fricking beautiful (see attachment)

3. It is just a great planet I don't think there needs to be any reason given I meannn you agree?

4. There's no other reason needed it's fricking gorgeous and amazing it needs no other reason. It needs a freaking

opening announcement to announce the arrival of the gorgeouness.

5 0
3 years ago
Read 2 more answers
Block B has mass 6.00 kg and sits at rest on a horizontal, frictionless surface. Block A has mass 2.50 kg and sits at rest on to
zzz [600]

Answer:

Explanation:

Block A sits on block B and force is applied on block A . Block A will experience two forces 1) force P and 2 )  friction force in opposite direction of motion . Block B will experience one force that is force of friction in the direction of motion .

Let force on block A be P . friction force on it will be equal to kinetic friction, that is μ mg , where μ is coefficient of friction and m is mass of block A

friction force = .4 x 2.5 x 9.8

= 9.8 N

net force on block A = P - 9.8

acceleration = ( P - 9.8 ) / 2.5

force on block B = 9.8

acceleration = force / mass

= 9.8 / 6

for common acceleration

( P - 9.8 ) / 2.5  = 9.8 / 6

( P - 9.8 ) / 2.5 = 1.63333

P = 13.88 N .

4 0
4 years ago
A child whirls a 3.00 kg ball on a string .50 m from the axis of rotation in a horizontal circle. The ball makes 1 revolution in
melamori03 [73]

Answers:

a) 0.5 m/s^{2}

b) 1.5 N

Explanation:

a) The centripetal acceleration a_{c} of an object moving in a uniform circular motion is given by the following equation:  

a_{c}=\omega^{2} r  

Where:

\omega=1 \frac{rev}{s} is the angular velocity of the ball

r=0.5 m is the radius of the circular motion, which is equal to the length of the string

Then:

a_{c}=(1 \frac{rev}{s})^{2} 0.5 m  

a_{c}=0.5 m/s^{2} This is the centripetal acceleration of the ball

b) On the other hand, in this circular motion there is a force (centripetal force F) that is directed towards the center and is equal to the tension (T) in the string:

F=T=m. a_{c}

Where m=3 kg is the mass of the ball

Hence:

T=(3 kg)(0.5 m/s^{2})

T=1.5 N This is the tension in the string

7 0
3 years ago
A rocket of mass 40000kg propelled by a force of 10^6N acquires a speed of 3000m/s determine the power expended
Katyanochek1 [597]

Answer:

3×10⁹ W

Explanation:

Power = work / time

Power = force × distance / time

Power = force × velocity

P = Fv

P = (10⁶ N) (3000 m/s)

P = 3×10⁹ W

6 0
3 years ago
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