All categories

3 years ago

How much force is required to accelerate a 5 kg mass at 20 m/s^2

Physics

1 answer:

Studentka2010 [4]3 years ago

6 0

Hello!

$\large\boxed{F = 100N}$

Use the equation F = m · a (Newton's Second Law) to solve. Substitute in the given values:

F = 5 · 20

F = 100N

You might be interested in

The wall is 500 sq. feet. A gallon of paint covers 160 sq. feet. What is an appropriate conversion factor to help determine how

lana66690 [7]

Answer:

Explanation:

Given

Wall is $500 ft^2$ in area

$160\ ft^2$ requires 1 gallon of paint

so using unitary method

$1\ ft^2 requires \frac{1}{160}$ gallon of paint

$500 ft^2$ wall will require $=500\times \frac{1}{160}$ gallons of paint

$=3.0125\ gallons$

4 0

3 years ago

Momentum data about the same objects in the same closed system is shown below.

nalin [4]

Answer:

the value of x is 3.7 because they are arranged in a particular manner.

6 0

4 years ago

Read 2 more answers

Tire or false

spayn [35]

1 and 4 are tire.
2 and 3 are not.

8 0

3 years ago

A cube icebox of side 3cm has a thickness of 5.0cm. If 4.0 kg of ice is put in the box estimate the amount of ice remaining afte

qaws [65]

Answer:

The amount of solid ice remaining after 6 hours is approximately 3.68664 kg

Explanation:

The given parameters are;

The side length of the cube box, s = 3(0) cm = 0.3 m

The thickness of the cube box, d = 5.0 cm = 0.05 m

The mass of ice in the box, m = 4.0 kg

The outside temperature of the cube box, T₁ = 45°C

The temperature of the melting ice inside the box, T₂ = 0°C

The latent heat of fusion of ice, $L_f$ = 3.35 × 10⁵ J/K/hr/kg

The surface area of the box, A = 6·s² 6 × (0.3 m)² = 0.54 m²

The coefficient of thermal conductivity, K = 0.01 J/s·m⁻¹·K⁻¹

For thermal equilibrium, we have;

The heat supplied by the surrounding = The heat gained by the ice

The heat supplied by the surrounding, Q = K·A·ΔT·t/d

Where;

ΔT = T₁ - T₂ = 45° C - 0° C = 45° C

ΔT = 45° C

Q = K·A·ΔT·t/d = 0.01 × 0.54 × 45 × 6× 60×60/0.05 = 104976

∴ The heat supplied by the surrounding, Q = 104976 J

The heat gained by the ice = $L_f$ × $m_{melted \ ice}$ =3.35 × 10⁵ J/kg × $m_{melted \ ice}$

Therefore, from Q = $L_f$ × $m_{melted \ ice}$ , we have;

Q = 104976 J = $L_f$ × $m_{melted \ ice}$ = 3.35 × 10⁵ J/kg × $m_{melted \ ice}$

104976 J = 3.35 × 10⁵ J/kg × $m_{melted \ ice}$

$m_{melted \ ice}$ = 104976 J/(3.35 × 10⁵ J/kg) ≈ 0.31336 kg

The mass of melted ice, $m_{melted \ ice}$ ≈ 0.31336 kg

∴ The amount of solid ice remaining after 6 hours, $m_{ice}$ = m - $m_{melted \ ice}$

Which gives;

$m_{ice}$ = m - $m_{melted \ ice}$ = 4.0 kg - 0.31336 kg ≈ 3.68664 kg

The amount of solid ice remaining after 6 hours, $m_{ice}$ ≈ 3.68664 kg.

8 0

3 years ago

A golfer hits a shot to a green that is elevated 3.20 m above the point where the ball is struck. The ball leaves the club at a

dem82 [27]

Answer:

16.17 m/s

Explanation:

h = 3.2 m

u = 18.1 m/s

Angle of projection, θ = 49°

Let H be the maximum height reached by the ball.

The formula for the maximum height is given by

$H=\frac{u^{2}Sin^{2}\theta }{2g}$

$H=\frac{18.1^{2}\times Sin^{2}49 }{2\times 9.8}=9.52 m$

The vertical distance fall down by the ball, h' H - h = 9.52 - 3.2 = 6.32 m

Let v be the velocity of ball with which it strikes the ground.

Use third equation of motion for vertical direction

$v_{y}^{2}=u_{y}^{2}+2gh'$

here, uy = 0

So,

$v_{y}^{2}=2\times 9.8 \times 6.32$

vy = 11.13 m/s

vx = u Cos 49 = 18.1 x 0.656 = 11.87 m/s

The resultant velocity is given by

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}$

$v=\sqrt{11.87^{2}+11.13^{2}}$

v = 16.27 m/s

7 0

4 years ago