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3 years ago

If an object is accelerating at 4 m/s^2 and has a mass of 8 kg, what is the net

Physics

2 answers:

Salsk061 [2.6K]3 years ago

7 0

Answer:

The force applied is 32 N

Explanation:

F = ma

F = 8 × 4

F = 32 N

cricket20 [7]3 years ago

4 0

Answer:

$\boxed {\boxed {\sf 32 \ Newtons}}$

Explanation:

We are asked to find the net force on an object. According to Newton's Second Law of Motion, force is the product of mass and acceleration. Therefore, the formula for calculating force is:

$F= m \times a$

The mass of the object is 8 kilograms and its acceleration is 4 meters per second squared.

m= 8 kg
a= 4 m/s²

Substitute these values into the formula.

$F= 8 \ kg \times 4 \ m/s^2$

Multiply.

$F= 32 \ kg *m/s^2$

Convert the units. 1 kilogram meter per second squared is equal to 1 Newton. So, our answer of 32 kilogram meters per second squared is equal to 32 Newtons.

$F= 32 \ N$

The net force on the object is <u>32 Newtons.</u>

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Light can travel from the sun to ________ in less than ten minutes.

Solnce55 [7]

To the Earth in less than ten minutes.

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3 years ago

A 10 m long uniform beam weighing 100 N is supported by two ropes at the ends. If a 400 N person sits at 2.0 m from one end of t

Naddik [55]

Answer:

T1 = 130N, T2 = 370N

Explanation:

In order for the system to be at rest, the sum of all forces must be zero and the torque around a point on the beam must be zero.

1. forces:

Let tension in rope 1 be T1 and in rope 2 be T2:

ma = T1 + T2 - 100N - 400N = 0

(1) T1 + T2 = 500N

2. torque around the center point of the beam:

τ = r x F = 5*T1 + 3*400N - 5*T2 = 0

(2) T1 - T2 = -240N

Solving both equations:

T1 = 130N

T2 = 370N

3 0

4 years ago

What is the difference between chemical change and a physical change?

NeX [460]

Answer:

Explanation:

There are several differences between a physical and chemical change in matter or substances. A physical change in a substance doesn't change what the substance is. In a chemical change where there is a chemical reaction, a new substance is formed and energy is either given off or absorbed.

3 0

3 years ago

A photographer in a helicopter ascending vertically at a constant rate of accidentally drops a camera out the window when the he

maksim [4K]

Answer:

The time it takes for the camera to reach the ground is 5 s.

Explanation:

To solve this problem, we will use the free fall cinematic equation.

Since the helicopter ascends with constant speed, the camera falls to the ground only by the effect of gravity on it.

The speed at which the helicopter ascends is not specified in the statement, but according to a similar problem, we will use 12.5 $\frac{m}{s}$ .

First, we must calculate the time and the maximum height at which the camera arrives after leaving the helicopter.

To calculate the maximum height to which it arrives, we will use the formula of vertical shot (since the camera leaves the helicopter with a speed upwards of 12,5 $\frac{m}{s}$ ).

$V_{f} ^{2}$ = $V_{0} ^{2}$ - 2 * g * h

Where:

$V_{f}$ : final speed at maximum height.

$V_{0}$ : initial speed when it falls from the helicopter.

g: gravity taken at 9.8 $\frac{m}{s^{2} }$

h: height reached from 60 m when leaving the helicopter

as $V_{f}$ =0

0= $(12,5\frac{m}{s}) ^{2}$ - 2 * $9,8\frac{m}{s^{2} }$ * h

clear h:

h= $(12,5 \frac{m}{s^{2} } )^{2}$ / (2 * $9,8 \frac{m}{s^{2} }$ )

h=7,97 m

Then we must calculate the time it takes to reach its maximum height:

$V_{f}$ = $V_{0}$ - g * t

t: time it takes to arrive from the moment it leaves the helicopter at its maximum height.

as $V_{f}$ =0

0=12.5 $\frac{m}{s}$ - 9.8 $\frac{m}{s^{2} }$ * t

clearing t

t=12.5 $\frac{m}{s}$ / 9.8 $\frac{m}{s^{2} }$

t=1.27 s.

Now we can calculate the time it takes to fall from the maximum height of 67.97 m.

The equation we will use is Y= $v_{0}*t+(\frac{g*t^{2} }{2} )$

where:

t: time it takes for the camera to fall.

Y: height from where the camera falls concerning the ground.

$v_{0}$ : initial speed of the camera at the time of starting the fall.

g: acceleration of gravity, estimated at 9.8 $\frac{m}{s^{2} }$

Step 1: As the helicopter ascends with constant speed, the initial speed of the camera at the moment of falling is 0.

$v_{0}$ =0

So the first term of our equation is nullified.

Step 2: To calculate the time it takes to fall, we clear "t" of the equation:

Y= $\frac{(g*t^{2})}{2}$

Y*2=(g* $t^{2}$ )

$\frac{Y*2}{g}$ = $t^{2}$

$\sqrt{\frac{Y*2}{g} }$ =t

Step 3: I replace the values with the incognites and get "t".

t= $\sqrt{\frac{67,97m*2}{9,8\frac{m}{s^{2} } } }$

t=3,73 s

The total time it takes for the camera to fall from the moment it leaves the helicopter is the sum of the time it takes to reach the maximum point of height and the time it takes to fall to the ground from that height.

t= 1,27 s + 3,73 s = 5 s

Have a nice day!

3 0

4 years ago

After a ball is thrown upward and is in the air, what will its acceleration do?

kifflom [539]

Acceleration will always be samei.e.

Acceleration=g= 9.81 m/s^2

8 0

3 years ago