A constant 20 N force is applied to a 7 kg box to push it along the ground. How

1. A ski-plane with a total mass of 1200 kg lands towards the west on a frozen lake at 30.0

igor_vitrenko [27]

Answer:

d = 229.5 m

Explanation:

It is given that,

Total mass of a ski-plane is 1200 kg

It lands towards the west on a frozen lake at 30.0 m/s.

The coefficient of kinetic friction between the skis and the ice is 0.200.

We need to find the distance covered by the plane before coming to rest. In this case,

$\mu mg=ma\\\\a=\mu g\\\\a=0.2\times 9.8\\\\a=1.96\ m/s^2$

It is decelerating, a = -1.96 m/s²

Now using the third equation of motion to find the distance covered by the plane such that :

$v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}\\\\d=\dfrac{-(30)^2}{2\times -1.96}\\\\d=229.59\ m$

So, the plane slide a distance of 229.5 m.

6 0

3 years ago

Find the resultant of the following forces: 3.0N east, 4.0N west, and 5.0N in a direction north 60° west.

zzz [600]

Answer:

5.5 N at 50.8° north of west.

Explanation:

To find the resultant of these forces, we have to resolve each force along the x- and y-direction, then find the components of the resultant force, and then calculate the resultant force.

The three forces are:

$F_1=3.0 N$ (east)

$F_2=4.0 N$ (west)

$F_3=5.0 N$ (at 60° north of west)

Taking east as positive x-direction and north as positive y-direction, the components of the forces along the 2 directions are:

$F_{1x}=3.0 N\\F_{1y}=0$

$F_{2x}=-4.0 N\\F_{2y}=0$

$F_{3x}=-(5.0)(cos 60^{\circ})=-2.5 N\\F_{3y}=(5.0)(sin 60^{\circ})=4.3 N$

Threfore, the components of the resultant force are:

$F_x=F_{1x}+F_{2x}+F_{3x}=3.0+(-4.0)+(-2.5)=-3.5 N\\F_y=F_{1y}+F_{2y}+F_{3y}=0+0+4.3=4.3 N$

Therefore, the magnitude of the resultant force is

$F=\sqrt{F_x^2+F_y^2}=\sqrt{(-3.5)^2+(4.3)^2}=5.5 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{|F_x|})=tan^{-1}(\frac{4.3}{3.5})=50.8^{\circ}$

And since the x-component is negative, it means that this angle is measured as north of west.

7 0

3 years ago

(a) Calculate the magnitude of the gravitational force exerted on a 445-kg satellite that is a distance of 1.77 earth radii from

victus00 [196]

Answer:

a)1396.52 N

b)1396.52 N

c) $a_{satellite}= 3.13 m/sec^2$

d) $a_{earth}=2.32\times10^{-22} m/s^2$

Explanation:

The force experienced by the satelite is giveb by

$F= \frac{Gm_{satellite}m_{earth}}{r^2} \\$

m_{satellite}= 445 Kg

m_{earth}= 6×10^24 Kg

radius r= 1.77Re= 1.77×6.38×10^6 m

now putting values we get

$F= \frac{6.67\times10^{-11}(445)(6\times10^24)}{(1.77\times6.38\times10^6)^2}$

⇒F= 1396.52 N

now,

$a_{satellite}= \frac{F}{m_{satellite}}$

$a_{satellite}= \frac{1396.52}{445}$

$a_{satellite}= 3.13 m/sec^2$

also,

$a_{earth}= \frac{F}{m_{earth}}$

$a_{earth}= \frac{1396.52}{(6\times10^24)}$

$a_{earth}=2.32\times10^{-22} m/s^2$

3 0

3 years ago

A microwave oven has 1200 W. If it receives 120 V of potential difference, what is the current in the microwave?

8_murik_8 [283]

Answer:

10amps

Explanation:

Given data

Power =1200W

Voltage =120V

Required

The current I

From

P=IV

I= P/V

Substitute

I=1200/120

I= 10amps

Hence the current is 10amps

7 0

2 years ago

A spherical balloon is being inflated and the radius of theballoon is increasing at a rate of 2 cm/s.(A) Express the radius (r)

goblinko [34]

Answer:

A.) r = 2t

B.) V = 33.5t^3

Explanation:

Given that a spherical balloon is being inflated and the radius of the balloon is increasing at a rate of 2 cm/s

A) Express the radius (r) of the balloon as a function of the time (t).

Since the rate = 2 cm/s that is,

Rate = radius/ time

Therefore,

2 = r/t

Make r the subject of formula

r = 2t

(B) If V is the volume of the balloon as a function of the radius, find V or and interpret it.

Let assume that the balloon is spherical. Volume of a sphere is;

V = 4/3πr^3

Substitute r = 2t into the formula

V = 4/3π(2t)^3

V = 4/3π × 8t^3

V = 32/3 × πt^3

V = 33.5t^3

6 0

3 years ago