What is the ratio of a 0.7 in circle in centimeters.

A 18.4-kg box rests on a frictionless ramp with a 16.1° slope. The mover pulls on a rope attached to the box to pull it up the i

laiz [17]

Answer:

F= 56,1 N :

Explanation:

We apply Newton's first law for balancing forces system.

We take the x axis in the direction of the ramp with a 16.1° slope.

∑Fx= 0

∑Fx: algebraic sum of forces ( + to the right, - to the left)

Problem development:

Look at the free body diagram :

The only forces that act on the box are the weight and tension of the rope because there is no friction.

T: Rope tension (N)

W :Box weight (N)

∑Fx= 0

Tx-Wx=0

Tx =Tcos( 43.2°- 16.1°)= Tcos ( 27.1°)

Wx= W*sen 16.1°= m*g*sen 16.1°=18.4*9.8* sen 16.1° = 50N

Tcos ( 27.1°)-50=0

Tcos ( 27.1°) = 50

T = (50) / (cos ( 27.1°))

T = 56,1 N

T=F= 56,1 N

3 0

3 years ago

A river has a steady speed of 0.550 m/s. A student swims upstream a distance of 1.00 km and swims back to the starting point. If

nlexa [21]

Answer:

$t=564.83seconds=9.414minutes=0.1569hours$

Explanation:

Here the steady speed of the river water relative to ground is

$V_{wg}=0.55m/s$

Speed of the boy relative to still water is

$V_{BW}=1.50m/s$

Here we are considering +ve speed along along +ve x-axis and -ve speed along -ve x-axis

Therefore the speed of boy relative to the ground upstream is:

$V_{BGup}=V_{BW}-V_{WG}\\V_{BGup}=1.50m/s-0.550m/s\\V_{BGup}=0.95m/s$

And the speed of boy relative to the ground downstream is:

$V_{BGdown}=V_{BW}+V_{WG}\\V_{BGdown}=1.50m/s+0.550m/s\\V_{BGdown}=2.05m/s$

The distance covered in upstream trip d₁=1.0km=1000m and the distance covered in downstream is d₂=1.0km=1000m

Since time taken by a person is to cover distance d with speed v given by:

$t=d/v$

The total time taken for one trip is:

$t=\frac{d_{1} }{V_{BGup} } -\frac{d_{2} }{V_{BGdown} }\\t=\frac{1000m}{0.95m/s}-\frac{1000m}{2.05m/s}\\ t=564.83seconds=9.414minutes=0.1569hours\\$

3 0

3 years ago

A car initially traveling at 21.4 m/s accelerates at a rate of 4.4 m/s2 for 7.5 seconds. What is the final velocity of the car?

Marysya12 [62]

The car's final velocity is 11.6 m/s in the opposite direction.

8 0

3 years ago

2. A marble is rolling at a velocity of 1.5 m/s with a momentum of 3.0 kg×m/s. What is its mass?

Nikolay [14]

Answer:

m= 2[kg]

Explanation:

Linear momentum can be calculated as the product of mass by Velocity.

$P=m*v$

where:

P = linear momentum = 3 [kg*m/s]

v = velocity = 1.5 [m/s]

m = mass [kg]

$m=P/v\\m=3/1.5\\m=2[kg]$

5 0

3 years ago

An electron is pushed into an electric field where it acquires a 1-v electric potential. if two electrons are pushed the same di

liq [111]

Given here that through the same region 2 electrons are pushed by same distance

As we know that electric potential can be relate with electric field by following relation

$\delta V = E . d$

Since the two electrons are pushed through same electric field by same distance

So the product of E and d will be same

So the electric potential will be same as initial

so V = 1 volt for two electrons

6 0

4 years ago