A proton is released from rest at x=5 cm. what direction does it move?

An airplane is flying in a horizontal circle at a speed of 100 m/s. The 80.0 kg pilot does not want the centripetal acceleration

Tema [17]

Answer:

The force will be "5,488". A further solution is provided below.

Explanation:

The given values are:

speed,

v = 100m/s

mass,

m = 80 kg

acceleration,

a = 7

Now,

Radius will be:

⇒ $r=\frac{v^2}{ac}$

⇒ $=\frac{(100)^2}{7\times 9.8}$

⇒ $=\frac{10000}{68.6}$

⇒ $=145.77$

Force will be:

⇒ $F=m\times ac$

⇒ $=80\times 68.6$

⇒ $=5488$

3 0

3 years ago

the velocity of a car increases from 2.0 m/s to 16.0 m/s in a time period of 3.5 s. what was the average acceleration? if an aut

Oksana_A [137]

The Average acceleration is 3.25m/s².

Acceleration is the change in velocity of the body with respect to time.

Where,

a= acceleration

u= initial speed

v= final speed

t= time

Initially, u = 2m/s

v = 16m/s

t = 3.5s

a = v-u/t

a = 16-2/3.5

a = 4m/s²

Then, u = 26m/s

v = 18m/s

t = 4s

a = v-u/t

a = 26-18/4

a =2 m/s²

Now, u = 0m/s

v = 20m/s

t = 2s

a = v-u/t

a = 20/2

a = 10m/s²

Finally, u = 20m/s

v = 10m/s

t = 2s

a = v-u/t

a = 10-20/2

a = -5m/s²

The average acceleration is the acceleration during the entire journey = 4+2+10-5/4

Average acceleration = 13/4m/s²

Average acceleration = 3.25m/s²

The Average acceleration is 3.25m/s²

Learn more about Acceleration here, brainly.com/question/2303856

#SPJ4

7 0

2 years ago

An archer shoots an arrow at an 85.0 m distant target; the bulls eye of the target is at same height as the release height of th

inessss [21]

Answer:

a) 14.1°

b) over

Explanation:

The usual model of ballistic motion assumes that the only force on the flying object is that due to gravity. When an object is launched with initial velocity v0 at some angle θ with respect to the horizontal, the distance it travels is ...

d = (v0)²sin(2θ)/g

Using this relation, we can find the launch angle to make the object travel a given distance:

θ = 1/2arcsin(dg/v0²) . . . . where g is the acceleration due to gravity

__

For the arrow to hit a target 85 m away at the same height it was launched with speed 42.0 m/s, the launch angle must be ...

θ = 1/2arcsin(dg/v0²) = 1/2(arcsin(85·9.8/42²)) ≈ 14.0893°

The arrow must be released at an angle of about 14.1°.

__

The flight time to the tree at a distance of 42.5 m will be that distance divided by the horizontal speed:

t = 42.5/(42cos(14.0893°)) ≈ 1.0433 . . . . seconds

The height at that time is ...

h(t) = -4.9t² +42sin(14.0893°)t ≈ 5.33 . . . meters

The arrow will go <em>over</em> the branch.

_____

<em>Additional comment</em>

Since gravity provides the only force on the arrow, its horizontal speed is constant at vh = v0·cos(θ), when the arrow is launched with speed v0 at angle θ above the horizontal. Its vertical speed will be reduced by the acceleration of gravity, so will be vv = v0·sin(θ) -gt. The height is the integral of the vertical speed, so is ...

h(t) = (1/2)gt² +v0·sin(θ)t

The height will be 0 at t=0 and at t=2v0sin(θ)/g, so the horizontal distance traveled will be ...

d = vh·t

= (v0·cos(θ))(2v0·sin(θ)/g) = (v0²/g)(2·sin(θ)cos(θ))

= v0²sin(2θ)/g

Note that this is all simplified by the fact that the target and launch point are at the same level (h=0).

6 0

3 years ago

Compare and contrast carbon to its isotopes: what is different about them and what is the same?

Alex_Xolod [135]

Their atomic numbers are all the same, because they all have the same number of protons. But their atomic weights are different, because they have different numbers of neutrons.

6 0

4 years ago

A firecracker is thrown downward from a height of 2.75m above the ground, with a speed of 3.15m/s. Ignore air resistance, determ

3241004551 [841]

Here in this question as we can see there is no air friction so we can use the principle of energy conservation

$PE_i + KE_i = PE_f + KE_f$