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3 years ago

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An object is moving at a velocity of 3m/s. It accelerates to a velocity of 5.5m/s over a time of 9 s. What acceleration did it e

xperience?

1 answer:

alexira [117]3 years ago

3 0

Answer:

Acceleration(a) = $\displaystyle\frac{v - u}{t}$

v = final velocity = 5.5 m/s

u = initial velocity = 3 m/s

t = time = 9s

$\displaystyle a = \frac{3-5.5}{9}$

$a = \displaystyle\frac{-2.5}{9}$

$a = -0.278 \ m/s^2$ (approx)

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An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the

zzz [600]

Answer:

The magnitude of the electric flux is $3.53\ N-m^2/C$

Explanation:

Given that,

Electric field = 2.35 V/m

Angle = 25.0°

Area $A= 1.65 m^2$

We need to calculate the flux

Using formula of the magnetic flux

$\phi=E\cdot A$

$\phi = EA\cos\theta$

Where,

A = area

E = electric field

Put the value into the formula

$\phi=2.35\times1.65\times\cos 25^{\circ}$

$\phi=2.35\times1.65\times0.91$

$\phi=3.53\ N-m^2/C$

Hence, The magnitude of the electric flux is $3.53\ N-m^2/C$

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3 years ago

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Alexxandr [17]

Explanation:

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3 years ago

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4 years ago

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What happens when your wearing socks and you shuffle across a carpet then touch a door knob?

natali 33 [55]

Answer:

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3 years ago

A particle's position along the x-axis is described by: x(t)=A t + B t2

ki77a [65]

Answer:

a)V = 25.1 m/s

b)V = 4.226 m/s

Explanation:

Given that

x(t)=A t + B t²

A = -4.3 m/s

B = 4.9 m/s²

x(t)= - 4.3 t +4.9 t²

The velocity of the particle is given as

$V=\dfrac{dx}{dt}$

V=-4.3 + 4.9 x 2 t

V= - 4.3 + 9.8 t m/s

Velocity at point t= 3 s

V= - 4. 3 + 9.8 x 3 m/s

V= - 4.3 + 29 .4 m/s

V = 25.1 m/s

At origin :

x= 0 m

0 = - 4.3 t +4.9 t²

0 = - 4.3 + 4.9 t

$t=\dfrac{4.3}{4.9}\ s$

t=0.87 s

The velocity at t= 0.87 s

V= - 4.3 + 9.8 t m/s

V= - 4. 3 + 9.8 x 0.87 m/s

V= - 4.3 + 8.526 m/s

V = 4.226 m/s

a)V = 25.1 m/s

b)V = 4.226 m/s

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3 years ago