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3 years ago

Which of these is an example of acceleration ?

Physics

2 answers:

anygoal [31]3 years ago

8 0

Answer:

no picture or anything so cant anwser

Explanation:

charle [14.2K]3 years ago

5 0

Answer:

can’t answer there’s no picture or options

Explanation:

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In a power plant, pipes transporting superheated vapor are very common. Superheated vapor flows at a rate of 0.3 kg/s inside a p

grigory [225]

Answer: $h=160.84 W/m^2-K$

Explanation:

Given

mass flow rate=0.3 kg/s

diameter of pipe=5 cm

length of pipe=10 m

Inside temperature=22

Pipe surface =100

Temperature drop=30

specific heat of vapor(c)=2190 J/kg.k

heat supplied $Q=mc\Delta T=0.3\times 2190\times (30)$

Heat due to convection =hA(100-30)

$A=\pi d\cdot L$

$A=\pi 0.05\times 10=1.571 m^2$

$Q_{convection}=h\times 1.571\times (100-22)=122.538 h$

$Q=Q_{convection}$

19,710=122.538 h

$h=160.84 W/m^2-K$

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2 years ago

You ride your bike north for 100 m at a constant speed of 5 m/Your acceleration is…

inessss [21]

Answer:

our acceleration is 0 because there is a constant speed..

a = final speed - initial speed / time

final speed = initial speed ( since speed is constant)

therefore acceleration is 0

5 0

2 years ago

A greyhound's velocity changes from rest to 19 m/s in 2 seconds. What is the greyhound's average acceleration?

Andrew [12]

It should be b)9.5m/s2

6 0

2 years ago

On the Moon, the acceleration due to the effect of gravity is only about 1/6 of that on Earth. An astronaut whose weight on Eart

vazorg [7]

And on the moon, the with is 595/6 N

3 0

3 years ago

A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential

castortr0y [4]

Explanation:

Given that,

Electric field = 5750 N/C

Charge $q=+10.5\times10^{-6}\ C$

Distance = 5.50 cm

(a). When the charge is moved in the positive x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot d$

$\Delta U=-q(E\cdot d)$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}$

$\Delta U=-3.32\times10^{-3}\ J$

The change in electric potential energy is $-3.32\times10^{-3}\ J$

(b). When the charge is moved in the negative x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot (-d)$

$\Delta U=-q(E\cdot (-d))$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})$

$\Delta U=3.32\times10^{-3}\ J$

The change in electric potential energy is $3.32\times10^{-3}\ J$

Hence, This is the required solution.

3 0

3 years ago