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3 years ago

5

A plane stops from 250 mph in 25 seconds. Calculate the plane's acceleration.

1 answer:

Lelechka [254]3 years ago

6 0

Explanation:

Given:

v₀ = 250 mph

v = 0 mph

t = 25 s

Find: a

v = at + v₀

(0 mph) = a (25 s) + (250 mph)

a = -10 mph/s

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A ball is moving with velocity 5 m/s in a direction which makes an angle of 30° with horizontal (i.e. with positive x-direction)

klasskru [66]

Answer:

Explanation:

(a)

From the given information:

The initial velocity $v_1$ = 5 m/s

The direction of the angle θ = 30°

Therefore, the component along the x-axis = $v_1 \ cos \ \theta$

$v_{1 \ x } = 5 \ cos \ 30^0$

$v_{1x} = 4.33 \ m/s$

The component along the y-axis = $v_2 { \ sin \ \theta}$

$v_{1 \ y } = 5 \ sin \ 30^0$

$v_{1 \ y } = 2.5 \ m/s$

To find the final velocity( reflected velocity)

using the same magnitude $v_2 = 5 \ m/s$

The angle from the x-axis can be $\theta_r = 90^0+60^0$

= 150°

Thus, the component along the x-axis = $v_2 \ cos \theta _r$

$v_{2x} = - 0.433 \ m/s$

The component along the y-axis = $v_2 \ sin \theta_r$

$v_{2y} = 5 \ sin \ 150^0$

$v_{2y} = 2.5 \ m/s$

(b)

The velocity $v_1$ can be written as in vector form.

$v_1 ^{\to} = v_1 x \hat {i} + v_1 y \hat {j}$

$v_1 ^{\to} =4.33 \ \hat {i} + 2.5 \ \hat {j}$ ---- (1)

The reflected velocity in vector form can be computed as:

$v_2 ^{\to} = v_2 x \hat {i} + v_2 y \hat {j}$

$v_2 ^{\to} =-4.33 \ \hat {i} + 2.5 \ \hat {j}$ --- (2)

The change in velocity = $v_2 ^{\to} - v_1 ^{\to}$

$\Delta v ^{\to} = - 4.33 \hat i + 2.5 \hat j - 4.33 \hat i - 2.5 \hat j$

$\Delta v ^{\to} = - 8.66 \hat { i }$

(c)

The magnitude of change in velocity = $| \Delta V |$

$| \Delta V |$ = 8.66 m/s

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3 years ago

If a seagull drops a shell from rest at a height of 12 m how fast is the shell moving when it hits the rocks

Dvinal [7]

As seagull drops a shell from rest at a height of 12 m, so we use kinematic equation of motion,

$v^{2} = u^{2} +2g h$

Here, h is the height, u is initial velocity , v is final velocity and g is acceleration due to gravity.

Given, h = 12 m.

We take, $g = 9.8 \ m/s^2$ and $u = 0$ because seagull drops a shell from rest.

Therefore, the speed of shell when it hits the rocks,

$v^{2} = 0 + 2 \times 9.8 m/s^2 \times 12 \ m = 235.2 (m/s)^2 \\\\ v = \sqrt{235.2 (m/s)^2} = 15.33 \ m/s^2$

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In photosynthesis, carbon dioxide is converted to oxygen and then released.

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What is the question yes it is converted to and then released

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Static discharges occur when

vredina [299]

The sudden flow of electricity between two electrically charged objects caused by contact, an electrical short, or dielectric breakdown. A buildup of static electricity can be caused by tribocharging or by electrostatic induction.

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Fifteen identical particles have various speeds: one has a speed of 2.00 m/s, two have speeds of 3.00 m/s, three have speeds of

Evgesh-ka [11]

Answer:

a) $V=7.5m/s$

b) $rms=8.4m/s$

c) Generally the most probable speed is 8m/s as it the most posses by particles being the average

Explanation:

From the question we are told that:

Sample size N=15

$Speed 1 v_1=2m/s\\\\Speed 2 v_2=3m/s\\\\Speed 3 v_2=5m/s\\\\Speed 4 v_4=8m/s\\\\Speed 3 v_5=9m/s\\\\Speed 2 v_6=15m/s\\\\$

Generally the equation for Average speed is mathematically given by

$V_{avg}=\frac{\sum(nv)}{N}$

Therefore

$V_{avg}=\frac{(2+2(3)+3(5)+4(8)+3(9)+2(15))}{15}$

$V=7.5m/s$

b)

Generally the equation for RMS speed of the particle is mathematically given by

$rms=\sqrt{\frac{\sum(nv^2)}{N}}$

$rms=\sqrt{\frac{2^2+2(3)^2+3(5)^2+4(8)^2+3(9)^2+2(15)^2}{15}}$

$rms=\sqrt{69.73}$

$rms=8.4m/s$

c

Generally the most probable speed is 8m/s as it the most posses by particles being the average

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3 years ago