Why is a device never 100% efficient? *

The possible presence of a supermassive black hole at the center of our galaxy has been deduced from

NNADVOKAT [17]

The movement of stars near the center.

8 0

3 years ago

Name the force between the brakes and the wheels that does work to slow down the vehicle when the brakes are applied

Feliz [49]

when brakes are applied then the tyres are stopped by brake- shoe

these brake shoe are rubber grip which hold tight at the rim of tyre

when these brake show hold tight at the rim then it will apply normal force on the tyre

Due to the rough surface of rubber and rim of the tyre these two surface apply force of friction on these two surfaces

Due to this friction force it resist the motion of tyre and due to which tyre will stop.

So the main cause to stop the tyre will be FRICTION force between brake shoe and tyre rim

7 0

4 years ago

A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra

Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

$sin(\alpha) = \frac{Vyi}{Vi}$

$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

$Vy(t) = Vyi -g.t$

where g is the gravity acceleration

$Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t$

This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

$Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}$

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

$Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s$

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

$Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} } .(0.675s)^{2} \\Y(0.675s) =2.236 m$

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

$Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m$

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0

4 years ago

Find the time it would take a car going 25 meters per seconds,east, to travel 400 meters per seconds,east

nexus9112 [7]

400÷25= 16seconds is the answer

3 0

3 years ago

A ball has a mass of 4.65kg and approximates a ping pong ball of mass 0.060kg that is at rest by striking it in an elastic colli

Alona [7]

Answer:

Look at work

Explanation:

Elastic Collision: Ki=Kf

M1=4.65kg

M2: 0.060kg

v1=5m/s

v2=0m/s

4.65*5+0.060*0=4.65*v1'+0.060*v2'

23.25+0=4.65v1'+0.060v2'

Also since it is an elastic collision we can use

v1+v1'=v2+v2'

4.65+v1'=v2'

Substitute into the earlier equation

23.25=4.65v1'+0.060(4.65+v1')

Expand

23.25=4.65v1'+0.279+0.06v1'

Solve for v1'

22.971=4.71v1'

v1'=4.88m/s

v2'=4.65+4.88=9.53m/s

7 0

3 years ago