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kari74 [83]
3 years ago
8

Average mass of a toy car(small, hot wheels)?

Physics
1 answer:
Tresset [83]3 years ago
6 0
2.4 ounces is the average weight of a hot wheels car.
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Click on the reset button, and stack one 50kg. crate on top of the other, so that the total mass is 100kg. The Friction should b
Helen [10]

Answer:

490.5 N

Explanation:

Coefficient of friction is 0.5 since friction force is set to halfway between none and lots. Minimum force is given by multiplying the weight and coefficient of friction

F= kN where k is coefficient of friction while N is weight. Also, N=mg where m is mass and g is acceleration due to gravity.

F=kmg=0.5*100*9.81=490.5 N

6 0
3 years ago
A man is 40 kg . calculate his weight.(g=10m/s^2)
Dafna1 [17]

{ \large{ \bf{Weight = mg}}}

{ \large{ \bf{Mass = 40 \:  kg \:  \:  \:  \: (given)}}}

{ \large{ \bf{Weight = 40 \times 10}}}

{ \large{ \therefore{ \bf{ \red{ Weight = 400N}}}}}

5 0
3 years ago
Can you explain the three main important reason why we use machine​
Maru [420]

Answer:

HERE IS YOUR ANSWER

Explanation:

1. The machines makes pur life easier.

2. It makes us do our works faster.

3. It does all the works that are harder for us.

Hope it helps you.

8 0
2 years ago
Read 2 more answers
What is the power of a parallel circuit with a resistance of 1,000 and a current of 0.03 A?
Kobotan [32]

i squared r = 0.03x0.03x1000=3x0.03x10=.9W

7 0
3 years ago
A motion sensor emits sound, and detects an echo 0.0115 s after. A short time later, it again emits a sound, and hears an echo a
Mekhanik [1.2K]

Answer:

1.17 m

Explanation:

From the question,

s₁ = vt₁/2................ Equation 1

Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.

Given: v = 343 m/s, t = 0.0115 s

Substitute into equation 1

s₁ = (343×0.0115)/2

s₁ = 1.97 m.

Similarly,

s₂ = vt₂/2.................. Equation 2

Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo

Given: v = 343 m/s, t₂ = 0.0183 s

Substitute into equation 2

s₂ = (343×0.0183)/2

s₂ = 3.14 m

The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁

s₂-s₁ =  (3.14-1.97) m = 1.17 m

7 0
3 years ago
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