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2 years ago

If a force of 10 N is applied on a mass of 5 kg, what will be its acceleration in m/s^2

Physics

1 answer:

Brut [27]2 years ago

6 0

Answer:

$a=2\ m/s^2$

Explanation:

<u>Second Newton's Law</u>

The theory that explains the relationship between the forces applied to an object and its motion is the second Newton's law.

It states that the net force applied to an object of mass m is given by

$F_n=m.a$

Where a is the acceleration aquired by the object that will cause a change of its velocity.

In our problem, we must compute the acceleration

$\displaystyle a=\frac{F_n}{m}=\frac{10}{5}$

$\boxed{a=2\ m/s^2}$

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the very high voltage needed to create a spark across the spark plug is produced at the a. transformer's primary winding. b. tra

Karo-lina-s [1.5K]

I think the correct answer from the choices listed above is option B. The very high voltage needed to create a spark across the spark plug is produced at the transformer's secondary winding. <span>The secondary coil is engulfed by a powerful and changing magnetic field. This field induces a current in the coils -- a very high-voltage current.</span>

5 0

2 years ago

Read 2 more answers

A 26 kg body is moving through space in the positive direction of an x axis with a speed of 350 m/s when, due to an internal exp

Gnom [1K]

Answer:

a.) 1567.2 m/s

b.) 149.4 m/s

Explanation:

Given that a 26 kg body is moving through space in the positive direction of an x axis with a speed of 350 m/s when, due to an internal explosion, it breaks into three parts. One part, with a mass of 7.8 kg, moves away from the point of explosion with a speed of 180 m/s in the positive y direction. A second part, with a mass of 8.8 kg, moves in the negative x direction with a speed of 640 m/s.

The x-component of the third part can be calculated by assuming that it moves in a positive x axis.

The third mass = 26 - ( 7.8 + 8.8)

The third mass = 26 - 16.6

The third mass = 9.4kg

since momentum is conserved, the momentum before explosion will be equal to sum of the momentum after explosion

26 x 350 = -8.8 x 640 + 9.4V

9100 = -5632 + 9.4V

9.4V = 9100 + 5632

9.4V = 14732

V = 14732/9.4

V = 1567.2 m/s

(b) y-component of the velocity of the third part will be

7.8 x 180 = 9.4 V

1404 = 9.4V

V = 1404/9.4

V = 149.4 m/s

7 0

2 years ago

A block of concrete has a mass of 48kg a crane lifts the block to a height of 12m above the ground calculate the gravitational p

Afina-wow [57]

Answer:

5760 J

Explanation:

From the question given above, the following data were obtained:

Mass of block = 48 kg

Height (h) = 12 m

Gravitational field strength (g) = 10 N/Kg

Gravitational potential energy (PE) =?

The gravitational potential energy stored by the block can simply be obtained as follow:

PE = mgh

PE = 48 × 10 × 12

PE = 5760 J

Therefore, the gravitational potential energy stored by the block is 5760 J

3 0

2 years ago

An object moving at 15 m/s slows uniformly at a rate of 2.0 m/s each second for 5.0 s. What is its final speed?

NeX [460]

Hey there!

We are given ,

Acceleration, a = -2m/s^2

Initial velocity , u = 15m/s

Time , t = 5 seconds

We know that ,
V=u+at

Now , final speed ,

V = 15+(-2)(5)

V = 15-10

V = 5 m/s -> final speed

Hope this helps you dear :)
Have a good day <3

8 0

1 year ago

The 4kg head of a sledge hammer is moving at 6m/s when it strikes a chisel, driving it into a log. The duration of the impact (o

LUCKY_DIMON [66]

Answer:

The average impact force is 12000 newtons.

Explanation:

By Impact Theorem we know that impact done by the sledge hammer on the chisel is equal to the change in the linear momentum of the former. The mathematical model that represents the situation is now described:

$\bar F \cdot \Delta t = m \cdot (v_{2}-v_{1})$ (1)

Where:

$\bar F$ - Average impact force, in newtons.

$\Delta t$ - Duration of the impact, in seconds.

$m$ - Mass of the sledge hammer, in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final velocity, in meters per second.

If we know that $\Delta t = 0.0020\,s$ , $m = 4\,kg$ , $v_{1} = -6\,\frac{m}{s}$ and $v_{2} = 0\,\frac{m}{s}$ , then we estimate the average impact force is:

$\bar F = \frac{m\cdot (v_{2}-v_{1})}{\Delta t}$

$\bar F = 12000\,N$

The average impact force is 12000 newtons.

5 0

2 years ago