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saw5 [17]
3 years ago
6

What is the acceleration of a 12kg box being pushed with 60N of force?

Physics
1 answer:
Mazyrski [523]3 years ago
5 0
60/12 =5m/s^2
Force/mass
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4 years ago
Determine the maximum theoretical speed that may be achieved over a distance of 66 m by a car starting from rest, knowing that t
My name is Ann [436]

Answer:

v=32.49 m/s

Explanation:

Given that

Distance ,d= 66 m

Initial speed of the car ,u = 0 m/s

Coefficient of friction ,μ = 0.8

Lets take the total mass of the car = m

The acceleration of the car is given as

a = μ g                               ( g= 10 m/s² )

Now by putting the values in the above equation we get

a= 0.8 x 10 m/s²

a= 8 m/s²

We know that ,final speed is given as

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Now putting the value

v²=0² + 2 x 8 x 66

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3 years ago
A scientist claimed that fabric A is better able to resist fire than fabric B. Which option describes an experiment that will pr
UNO [17]

Answer:

B. Hold each type of fabric over a candle flame and time how long it takes for the fabric to start to burn.

Explanation:

6 0
4 years ago
Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a –80-nC charge at x = –4 m on
leva [86]

Answer:48 V

Explanation:

Given

Three charged particle with charge

q_1=50\ nC at y=6\ m

q_2=-80\ nC at x=-4\ m

q_3=70\ nC at y=-6\ m

Electric Potential is given by

V=\frac{kQ}{r}

Distance of q_1 from x=8\ m

d_1=\sqrt{6^2+8^2}

d_1=\sqrt{36+64}

d_1=10\ m

similarly d_2=8-(-4)

d_2=12\ m

d_3=\sqrt{(-6)^2+8^2}

d_3=\sqrt{36+64}

d_3=10\ m

Potential at x=8\ m is

V_{net}=\frac{kq_1}{d_1}+\frac{kq_2}{d_2}+\frac{kq_3}{d_3}

V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]

V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}

V_{net}=9\times 5.33

V_{net}=47.97\approx 48\ V

5 0
3 years ago
Examples of uniform speed​
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A body is said to be moving with uniform speed, if it covers equal distances in equal intervals of time. ...

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