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3 years ago

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WallyGPX accelerates from 0 m/s to 8 m/s in 3 seconds. What is his acceleration? Is this acceleration higher than that of a car

which accelerates from 0 to 30 m/s in 8 seconds?

1 answer:

olga nikolaevna [1]3 years ago

4 0

My Phone is +2348181686682

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Un avión tarda en llegar a su destino 12 horas. Si recorrió una distancia de 10.700 kilómetros. Calcular su velocidad y expesarl

elixir [45]

Given that,

Distance, d = 10700 km

Time taken by the airplane to complete the destination, t = 12 hours

We need to find the speed of the airplane. It is equal to the total distance covered divided by total time taken. So,

$v=\dfrac{d}{t}\\\\v=\dfrac{10700\ km}{12\ h}\\\\v=891.66\ km/h$

We know that,

1 km = 1000 m

1 h = 3600 s

So,

$v=891.66\ km/h =\dfrac{891.66\times 1000\ m}{3600\ s}\\\\v=247.68\ m/s$

So, the speed of the airplane is 891.66 km/h or 247.68 m/s.

5 0

3 years ago

Jason and Guy are throwing watermelons straight up from the ground. (They’re making fruit salad.) Jason throws his watermelon at

NISA [10]

Answer:2t

Explanation:

Given

Jason throwing watermelon at speed v and it spend t time in air

time to reach max height

$v_f=u_i+at'$

0=v-gt'

$t'=\frac{v}{g}$

and time to reach bottom is also t'

t=2t'

$t=2\frac{v}{g}$

For guy throws his watermelon at speed 2v

So total time in air will be

$t''=2\frac{2v}{g}$

$t''=4\frac{v}{g}$

t''=2t

5 0

3 years ago

HELLLP PLEASE || the graph below shows a conversion of energy for a skydive jumping out of a plane and landing safely on the gro

fenix001 [56]

Answer: I maybe wrong but i'm pretty sure its C) Kinetic energy

5 0

3 years ago

The potential at location A is 382 V. A positively charged particle is released there from rest and arrives at location B with a

jarptica [38.1K]

Answer: 247.67 V

Explanation:

Given

Potential At A $V_a=382\ V$

Potential at $V_c=785\ V$

when particle starts from A it reaches with velocity $v_b$ at Point while when it starts from C it reaches at point B with velocity $2v_b$

Suppose m is the mass of Particle

Change in Kinetic Energy of particle moving under the Potential From A to B

$q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1$

Change in Kinetic Energy of particle moving under the Potential From C to B

$q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2$

Divide 1 and 2 we get

$\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}$

on solving we get

$V_b=\frac{4}{3}\cdot V_a-\frac{1}{3}\cdot V_c$

$V_b=\frac{743}{3}=247.67\ V$

4 0

3 years ago

The amount of energy that must be absorbed or lost to raise or lower the temperature of 1 g of liquid water by 1°c _____.

Fiesta28 [93]

Answer:

4.2 J

Explanation:

Specific heat capacity: This is defined as the amount of a heat required to rise a unit mass of a substance through a temperature of 1 K

From specific heat capacity,

Q = cmΔt.............................. Equation 1

Where Q = amount of energy absorbed or lost, c = specific heat capacity of water, m = mass of water, Δt = Temperature rise.

Given: m = 1 g = 0.001 kg, Δt = 1 °C

Constant : c = 4200 J/kg.°C

Substitute into equation 1

Q = 0.001×4200(1)

Q = 4.2 J.

Hence the energy absorbed or lost = 4.2 J

6 0

3 years ago