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3 years ago

An unbalanced force of 500 N is applied to a 75 kg object. What is the acceleration of the object?

Physics

2 answers:

Lubov Fominskaja [6]3 years ago

8 0

Answer:

Acceleration, $a=6.66\ m/s^2$

Explanation:

Given that,

Force acting on the object, F = 500 N

Mass of the object, m = 75 kg

We need to find the acceleration of the object. The force acting on the object is given by :

$a=\dfrac{F}{m}$

$a=\dfrac{500\ N}{75\ kg}$

$a=6.66\ m/s^2$

So, the acceleration of the object is $6.66\ m/s^2$ . Hence, this is the required solution.

NemiM [27]3 years ago

7 0

1. Define Newtons second law of motion (this will help put things into perspective)

2.Get the mass of the object (in this case 75 kg)

3.The net force acting on the object...find it (in this case 500 N)

4.Change the equation to F=ma (500=75a)

5.Divide both sides by 75 and that is the acceleration.

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In the Bohr model of the atom, an electron in an orbit has a fixed ____.

worty [1.4K]

The answer is C. an electron in an orbit has a fixed energy.

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3 years ago

A 117 kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 62.

Ivenika [448]

Answer:

I_syst = 278.41477 kg.m²

Explanation:

Mass of platform; m1 = 117 kg

Radius; r = 1.61 m

Moment of inertia here is;

I1 = m1•r²/2

I1 = 117 × 1.61²/2

I1 = 151.63785 kg.m²

Mass of person; m2 = 62.5 kg

Distance of person from centre; r = 1.05 m

Moment of inertia here is;

I2 = m2•r²

I2 = 62.5 × 1.05²

I2 = 68.90625 kg.m²

Mass of dog; m3 = 28.3 kg

Distance of Dog from centre; r = 1.43 m

I3 = 28.3 × 1.43²

I3 = 57.87067 kg.m²

Thus,moment of inertia of the system;

I_syst = I1 + I2 + I3

I_syst = 151.63785 + 68.90625 + 57.87067

I_syst = 278.41477 kg.m²

8 0

3 years ago

The position of an electron is given by , with t in seconds and in meters. At t = 3.99 s, what are (a) the x-component, (b) the

egoroff_w [7]

Answer:

A. Vx = 3.63 m/s

B. Vy = -45.73 m/s

C. |V| = 45.87 m/s

D. θ = -85.46°

Explanation:

Given that position, r, is given as:

r = 3.63tˆi − 5.73t^2ˆj + 8.16ˆk

Velocity is the derivative of position, r:

V = dr/dt = 3.63 - 11.46t^j

A. x component of velocity, Vx = 3.63 m/s

B. y component of velocity, Vy = -11.46t

t = 3.99 secs,

Vy = - 11.46 * 3.99 = -45.73 m/s

C. Magnitude of velocity, |V| = √[(-45.73)² + 3.63²]

|V| = √(2091.2329 + 13.1769)

|V| = √(2104.4098)

|V| = 45.87 m/s

D. Angle of the velocity relative to the x axis, θ is given as:

tanθ = Vy/Vx

tanθ = -45.73/3.63

tanθ = -12.6

θ = -85.46°

7 0

3 years ago

The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of

motikmotik

Explanation:

It is given that, the force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the car and the square of the car's speed such that,

$F\propto \dfrac{mgv^2}{r}$

$F=\dfrac{kmgv^2}{r}$

mg is the weight of the car

r is the radius of the curve

v is the speed of the car

Case 1.

F = 640 pounds

Weight of the car, W = mg = 2600 pound

Radius of the curve, r = 650 ft

Speed of the car, v = 40 mph

$640=\dfrac{k(2600)(40)^2}{650}$

k = 0.1

Case 2.

Radius of the curve, r = 750 ft

Speed of the car, v = 30 mph

$F=\dfrac{0.1\times 2600\times (30)^2}{750}$

F = 312 N

Hence, this is the required solution.

6 0

3 years ago

A defibrillator is used during a heart attack to restore the heart to its normal beating pattern. A defibrillator passes 18 A of

miskamm [114]

Answer:

q = 0.036 C

Explanation:

Given that,

Current passes through a defibrillator, I = 18 A

Time, t = 2 ms

We need to find the charge moved during this time. We know that,

Electric current = charge/time

$q=It$

Put all the values,

$q=18\times 0.002\\\\q=0.036\ C$

So, 0.036 C of charge moves during this time.

8 0

3 years ago