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3 years ago

How can a cyclist's acceleration change even if its velocity remains constant?

Physics

1 answer:

aleksley [76]3 years ago

5 0

It can't. Acceleration IS a change in velocity.

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Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at

Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

$v = \dfrac{2\pi R}{T}$

$v = \dfrac{2\pi\times 8}{4.30}$

v = 11.69 m/s

now, Force does the ring push on her at the top

$- N - m g = \dfrac{-mv^2}{R}$

$N + m g = \dfrac{mv^2}{R}$

$N = \dfrac{mv^2}{R}- m g$

$N = m(\dfrac{v^2}{R}- g)$

$N = 58\times (\dfrac{11.69^2}{8}- 9.8)$

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

6 0

2 years ago

What is the resistance at 20°C of a 2.0-meter length of tungsten wire with a cross-sectional area of 7.9 10^-7

Bad White [126]

Answer:

1.4 * 10 ^-1 Ω

Explanation:

Hi,

For this question, we gotta use the formula

R = pL/A

p = The resistivity of your material at 20°C

L = length of the wire

A = cross-sectional area

The resistivity of tungsten is 5.60 * 10^-8 at 20°C

By plugging the values, we get:

R = (5.60 * 10^-8)(2.0)/(7.9*10^-7) = 1.4 * 10 ^-1 Ω

8 0

2 years ago

Tina's calculations of the tarantula found that the spider was able to cover 20 centimeters in five seconds what was the average

Sergeu [11.5K]

<span>(20 cm)/(5 sec) = (0.20 meters)/(5 seconds)
</span>

8 0

3 years ago

A cubical Gaussian surface surrounds two positive charges, each has a charge q 1 1 = + 3.90 × 10 − 12 3.90×10−12 C, and three ne

Masteriza [31]

Answer:

The electric flux is zero because charge is zero.

Explanation:

Given that,

Positive charge $q_{1}=3.90\times10^{-12}\ C$

Negative charge $q_{2}=-2.60\times10^{-12}\ C$

We need to calculate the total charged

Using formula of charge

$Q_{enc}=2q_{1}+3q_{2}$

Put the value into the formula

$Q_{enc}=2\times3.90\times10^{-12}+3\times(-2.60\times10^{-12})$

$Q_{enc}=0$

We need to calculate the electric flux

Using formula of electric flux

$\phi=\dfrac{Q_{enc}}{\epsilon_{0}}$

Put the value into the formula

$\phi=\dfrac{0}{8.85\times10^{-12}}$

Hence, The electric flux is zero because charge is zero.

7 0

2 years ago

An object with a mass of 78kg is lifted through a height of 6 meters. How much work is done?

g100num [7]

Work = force x distance.

force = mass x acceleration

work = mass x acceleration x diastance

use acceleration of gravity in this problem

W (J) = m (kg) x a (m/s/s) x d (m)
W = 78 x 9.8 x 6
W = 4586.4

3 0

2 years ago