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3 years ago

I really need some with help with this anyone who can help with acceleration and velocity pleasee and thank you

Physics

1 answer:

kykrilka [37]3 years ago

3 0

Answer:

velocity equals distance divided by time and the unit you use after you get the answer is m/s

Explanation:

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True or false The cardiovascular system includes the heart and blood vessels.

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The answer to this question is false

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A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed pa

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Answer

given,

current (I) = 16 mA

circumference of the circular loop (S)= 1.90 m

Magnetic field (B)= 0.790 T

S = 2 π r

1.9 = 2 π r

r = 0.3024 m

a) magnetic moment of loop

M= I A

M= $16 \times 10^{-3} \times \pi \times r^2$

M= $16 \times 10^{-3} \times \pi \times 0.3024^2$

M= $4.59 \times 10^{-3}\ A m^2$

b) torque exerted in the loop

$\tau = M\ B$

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3 years ago

Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate fr

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Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = $\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

W = $F_{total} .d$

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = $\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

$F_{total} .d =\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

$F_{total} = \frac{\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}}{d}$

$F_{total=} \frac{\frac{1}{2} X 62 X6^{2} -\frac{1}{2} X 62 X2^{2} }{25}$

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

$F_{sprinter} = F_{total} + F_{wind} = 39.7 + 30 = 69.68 N$

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3 years ago

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A charged cloud system produces an electric field in the air near earth's surface. a particle of charge -2.0 × 10-9 c is acted o

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Part a)

Magnitude of electric field is given by force per unit charge

$E = \frac{F}{q}$

$E = \frac{4.3 * 10^{-6}}{2 * 10^{-9}}$

$E = 2150 N/C$

Part b)

Electrostatic force on the proton is given as

F = qE

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$F = 3.44 * 10^{-16} N$

PART C)

Gravitational force is given by

$F_g = mg$

$F_g = 1.6 * 10^{-27}*9.8$

$F_g = 1.57 * 10^{-26} N$

PART d)

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$\frac{F_e}{F_g} = 2.2 * 10^{10}$

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Answer:

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Explanation:

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