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2 years ago

A runner runs 300 m at an average speed of 3.0 m/s. She then runs another 300m at an average

Physics

1 answer:

Kaylis [27]2 years ago

3 0

Answer:

B. 4 m/s

Explanation:

v=d/t

Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.

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An airplane is delivering food to a small island. It flies 100 m above the ground at a speed of 150 m/s .

miss Akunina [59]

Answer:

The airplane should release the parcel $6.7*10^2$ m before reaching the island

Explanation:

The height of the plane is $y_0=100m$ , and its speed is v=150 m/s

When an object moves horizontally in free air (no friction), the equation for the y measured with respect to ground is

$y=y_0 - \frac{gt^2}{2}$ [1]

And the distance X is

x = V.t [2]

Being t the time elapsed since the release of the parcel

If we isolate t from the equation [1] and replace it in equation [2] we get

$X = V . \sqrt{\frac{2y_0}{g}}$

Using the given values:

$x = 150 m/s \sqrt{\frac{2\times 100m}{9.8 m/sec^2}}$

x = $6.7*10^2$ m

4 0

2 years ago

A body is projected upward at an angle of 30 degree to the horizontal at an initial speed of 200ms-.In how many seconds will it

Crazy boy [7]

Answer:

20.41 s

3534.80 m

Explanation:

In how many seconds will it reach the ground?

We are given the initial velocity of the body, which is 200 m/s at a 30° angle.

We know the acceleration in the vertical direction is -9.8 m/s², assuming that the upwards/right direction is positive and the downwards/left direction is negative.

Since we are using acceleration in the y-direction, let's use the vertical component of the initial velocity.

200 · sin(30) m/s

Let's use the fact that at the top of its trajectory, the body will have a final velocity of 0 m/s.

Now we have one missing variable that we are trying to solve for: time t.

Find the constant acceleration equation that contains v₀, v, a, and t.

v = v₀ + at

Substitute known values into the equation.

0 = 200 · sin(30) + (-9.8)t
-200 · sin(30) = -9.8t
t = 10.20408163

Recall that this is only half of the body's trajectory, so we need to double the time value we found to find the total time the body is in the air.

2t = 20.40816327

The body will reach the ground in 20.41 seconds.

How far from the point of projection would it strike?

We want to find the displacement in the x-direction for the body.

Let's find the constant acceleration equation that contains time t, that we just found, and displacement (Δx).

Δx = v₀t + 1/2at²

Substitute known values into the equation. Remember that we want to use the horizontal component of the initial velocity and that the acceleration in the x-direction is 0 m/s².

Δx = (200 · cos(30) · 20.40816327) + 1/2(0)(20.40816327)²
Δx = 3534.797567

The body will strike 3534.80 m from the point of projection.

4 0

1 year ago

A racecar driver on a flat racetrack steps on the gas, changing her velocity from 10 m/sec to 30 m/sec in 5 seconds. What is the

erastova [34]

Answer:

4m/s2

Explanation:

The following data were obtained from the question:

U (initial velocity) = 10m/s

V (final velocity) = 30m/s

t (time) = 5secs

a (acceleration) =?

Acceleration is the rate of change of velocity with time. It is represented mathematically as:

a = (V - U)/t

Now, with this equation i.e

a = (V - U)/t, we can calculate the acceleration of the race car as follow:

a = (V - U)/t

a = (30 - 10)/5

a = 20/5

a = 4m/s2

Therefore, the acceleration of the race car is 4m/s2

8 0

2 years ago

Read 2 more answers

A 60-μC charge is held fixed at the origin and a −20-μC charge is held fixed on the x axis at a point x = 1.0 m. If a 10-μC char

Aleksandr [31]

Answer:

Ek = 8,79 [J]

Explanation:

We are going to solve this problem, using the energy conservation principle

State 1 or initial state (charges at rest t=0)

E₁ = Ek + U₁

As charge are at rest Ek = 0

And U₁ has two components

U₁₂ = K * Q₁*Q₂ / 0,4 and U₃₂ = K*Q₃*Q₂ / 0,6

U₁₂ = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,4 ⇒ U₁₂ = 9*60*10*10⁻³/0,4

U₃₂ = - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,6 ⇒ U₃₂ = - 9*20*10*10⁻³/0,6

U₁₂ = 540*10⁻2/0,4 [J] ⇒13,5 [J]

U₃₂ = - 180*10⁻² /0,6 [J] ⇒ - 3 [J]

Then E₁ = E₁₂ + E₃₂

E₁ = 10,5 [J]

At the moment of Q₂ passing x = 40 cm or 0,4 m

E₂ = Ek + U₂

We can calculate the components of U₂ in this new configuration

U₂ = U₁₂ + U₃₂

U₁₂ = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,7 ⇒ U₁₂ = 9*60*10*10⁻³/0,7

U₁₂ = 540*10⁻²/0,7 U₁₂ = 7,71 [J]

U₃₂ = - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,3 ⇒ U₃₂ = - 9*20*10*10⁻³/0,3

U₃₂ = - 9*20*10⁻²/0,3

U₃₂ = - 6

U₂ = 7,71 -6

U₂ = 1,71 [J]

Then as

E₂ = Ek + U₂ and E₂ = E₁

Then

Ek + U₂ = E₁

Ek = 10,5 - U₂

Ek = 10,5 - 1,71

Ek = 8,79 [J]

5 0

1 year ago

A hot air balloon of total mass M (including passengers and luggage) is moving with a downward acceleration of magnitude a. As i

LUCKY_DIMON [66]

Answer:

The fraction of mass that was thrown out is calculated by the following Formula:

M - m = (3a/2)/(g²- (a²/2) - (ag/2))

Explanation:

We know that Force on a moving object is equal to the product of its mass and acceleration given as:

F = ma

And there is gravitational force always acting on an object in the downward direction which is equal to g = 9.8 ms⁻²

Here as a convention we will use positive sign with acceleration to represent downward acceleration and negative sign with acceleration represent upward acceleration.

Case 1:

Hot balloon of mass = M

acceleration = a

Upward force due to hot air = F = constant

Gravitational force downwards = Mg

Net force on balloon is given as:

Ma = Gravitational force - Upward Force

Ma = Mg - F (balloon is moving downwards so Mg > F)

F = Mg - Ma

F = M (g-a)

M = F/(g-a)

Case 2:

After the ballast has thrown out,the new mass is m. The new acceleration is -a/2 in the upward direction:

Net Force is given as:

-m(a/2) = mg - F (Balloon is moving upwards so F > mg)

F = mg + m(a/2)

F = m(g + (a/2))

m = F/(g + (a/2))

Calculating the fraction of the initial mass dropped:

$M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]$

5 0

2 years ago