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2 years ago

12

What is the height of a building that an object is dropped from if it has a mass of 3 kg and hits the ground with a velocity of

50 m/s?

1 answer:

Marizza181 [45]2 years ago

8 0

Answer:

125 m

Explanation:

m = 3kg

v = 50m/s

u = 0m/s

a = +g = 10m/s²

s = H = ?

using the formula,

v² = u² + 2as

50² = 0² + 2(10)(H)

2500 = 20H

H = 2500/20

H = 125m

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Express 50,000 m/s2 in scientific notation. 5.0 x 103m/s2 0.5 x 105m/s2 5.0 x 104m/s2

Tresset [83]

Answer would be 5.0x104m/s2

8 0

3 years ago

Read 2 more answers

(b) Fig. 1.1 shows two airports A and C. north SE с sea land क WE not to scale Fig. 1.1 An aircraft flies due north from A for a

GalinKa [24]

The addition of vectors and the uniform motion allows to find the answers for the questions about distance and time are:

The distance to go between airports A and C is 373.6 10³ m

The time to go from airport A to B is 2117 s

Vectors are quantities that have modulus and direction, so their addition must be done using vector algebra.

In this case the plane flies towards the North a distance of y = 360 10³ m at an average speed of v = 170 m / s, when arriving at airport B it turns towards the East and travels from x = 100 10³ m, until' it the distance reaches the airport C

Let's use the Pythagoras theorem to find the distance traveled

R = Ra x² + y²

R = 10³

R = 373.6 10³ m

They indicate the average speed for which we can use the uniform motion ratio

v = $\frac{\Delta y }{t}$

t = $\frac{\Delta y}{v}$

They ask for the time in in from airport A to B, we calculate

t = 360 10 ^ 3/170

t = 2.117 10³ s

In conclusion we use the addition of vectors the uniform motion we can find the answer for the question of distance and time are:

The distance to go between airports A and C B is 373.6 10³ m

The time to go from airport A to B is 2117 s

Learn more here: brainly.com/question/15074838

5 0

2 years ago

A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn

elena-s [515]

Answer:

The change in potential energy is $\Delta PE = - 3.8*10^{-16} \ J$

Explanation:

From the question we are told that

The magnitude of the uniform electric field is $E = 950 \ N/C$

The distance traveled by the electron is $x = 2.50 \ m$

Generally the force on this electron is mathematically represented as

$F = qE$

Where F is the force and q is the charge on the electron which is a constant value of $q = 1.60*10^{-19} \ C$

Thus

$F = 950 * 1.60 **10^{-19}$

$F = 1.52 *10^{-16} \ N$

Generally the work energy theorem can be mathematically represented as

$W = \Delta KE$

Where W is the workdone on the electron by the Electric field and $\Delta KE$ is the change in kinetic energy

Also workdone on the electron can also be represented as

$W = F* x *cos( \theta )$

Where $\theta = 0 ^o$ considering that the movement of the electron is along the x-axis

So

$\Delta KE = F * x cos (0)$

substituting values

$\Delta KE = 1.52 *10^{-16} * 2.50 cos (0)$

$\Delta KE = 3.8*10^{-16} J$

Now From the law of energy conservation

$\Delta PE = - \Delta KE$

Where $\Delta PE$ is the change in potential energy

Thus

$\Delta PE = - 3.8*10^{-16} \ J$

7 0

3 years ago

The froghopper Philaenus spumarius is supposedly the best jumper in the animal kingdom. To start a jump, this insect can acceler

Kay [80]

Answer:

Part a)

$v_f = 4 m/s$

Part b)

$t = 0.001 s$

Part c)

$d = 0.815 m$

Explanation:

Part a)

As we know that initially the grass hopper is at rest at the ground position

Now the acceleration is given as

$a = 4000 m/s^2$

distance of the legs that it stretched is given as

$s = 2.0 mm$

so we have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(4000)(0.002)$

$v_f = 4 m/s$

Part b)

time taken to reach this speed is given as

$v_f - v_i = at$

$4 - 0 = 4000 t$

$t = 0.001 s$

Part c)

as the grass hopper reach the maximum height its final speed would be zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 4^2 = 2(-9.81) d$

$d = 0.815 m$

5 0

3 years ago

Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?

frosja888 [35]

Answer:

The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

Explanation:

From the question given above, the following data were obtained:

Height to which the target is located = 50 m

Initial velocity (u) = 20 m/s

To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:

Initial velocity (u) = 20 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 20² – (2 × 10 × h)

0 = 400 – 20h

Collect like terms

0 – 400 = – 20h

– 400 = – 20h

Divide both side by – 20

h = – 400 / – 20

h = 20 m

Thus, the the maximum height to which the cannon ball attained is 20 m.

From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

3 0

2 years ago