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2 years ago

What is the acceleration of a 10 kg mass pushed by a 5N force?

Physics

1 answer:

ehidna [41]2 years ago

4 0

Answer:

0.5

Explanation:

F = ma

5 = 10 a

a = 5/10

a = 0.5

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A softball player moving 3.89 m/s

Ahat [919]

Answer:

0.119 s

Explanation:

Given that

$U=3.89\ m/s\\a=-1.44\ m/s^2\\S=4.8\ m$

Lets take final speed of the softball after covering 4.8 m = V

We know that

$V^2=U^2+2aS\\V^2=3.89^2-2\times 1.44\times 4.8\\V=3.7\ m/s$

Also We know that

$V=U+at\\$

Putting the value of V ,U and a in the previous equation We get

$3.7=3.89-1.44\times t\\t=0.119\ s$

Therefore slide time will be 0.119 s

3 0

3 years ago

What is happening in the diagram?

pishuonlain [190]

In the diagram, the ship send sound(?) waves to the water, to determine if there is anything there. If there is something like a sunken ship shown in the diagram, the waves return in a shorter time hence you can understand if theres something or now. This is the principle of radars and sonars.

3 0

3 years ago

Read 2 more answers

What is the name of the ratio of the voltage applied to a circuit and the current in a circuit

malfutka [58]

Resistance is the name of the ratio of the voltage applied to a circuit and the current in a circuit. Goes under <span>Ohm's Law</span>

4 0

3 years ago

A box weighing 12,000 N is parked on a 36° slope. What will be the component of the weight parallel to the plane that balances f

sweet-ann [11.9K]

Given :

A box weighing 12,000 N is parked on a 36° slope.

To Find :

What will be the component of the weight parallel to the plane that balances friction.

Solution :

The component of that will be parallel to the plane to balance friction is :

$F \ cos ( 90 - 36)^o\\\\F\ ( sin\ 36^o)$

Therefore, component of force to balance friction is F sin 36° .

Hence, this is the required solution.

5 0

2 years ago

In the diagram, q1=+6.25 * 10^ -8 C. What is the potential difference when you go from point A to point B? Include the correct s

Nimfa-mama [501]

Answer:

Moving a unit "positive" test charge from A to B will result in a reduction in potential

V = K Q / R potential at a point

V2 - V1 = K Q (1 / .4 - 1 / .15) = = k Q (.15 - .4) / .06 = -4.17 K Q

V2 - V1 = -4.17 * 9 & 10E9 * 6.25 E-8

V2 - V1 = -4.17 * 562.5 J/C

V = - 2346 Volts

7 0

2 years ago