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3 years ago

Kaya is riding her dirt bike eastward on a dirt road. She spots a pothole ahead.

Physics

2 answers:

Ratling [72]3 years ago

7 0

Answer:

Kaya's acceleration is 1.41666667.

Explanation:

To find the acceleration we need to subtract the starting speed to the slowed speed. So 14 - 5.5 = 8.5.

Now we divide 8.5 by the time, 6 seconds.

Our final answer is 1.41666667, or rounded, 1.42.

Citrus2011 [14]3 years ago

3 0

Answer: The acceleration of Kaya is .

Explanation:

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A 400 g ball swings in a vertical cirde at the end of

velikii [3]

Answer:

15.10m/s

Explanation:

The mass of the ball(m)=400g = 0.4kg

The radius of the string is(r)=15m

The tension in the string is(T)=10N

The acceleration due to gravity = $9.8m/s^{2}$

The tension in the string when the body is at the bottom is given by

$T=\frac{mv^{2} }{r}+mg$

To find the speed of the ball, we make v the subject of the formula

Therefore, $v=\sqrt\frac{r(T-mg}{m}$

$v= \sqrt\frac{15(10-0.4*9.8)}{0.4}$

$v=\sqrt\frac{91.2}{0.4} \\$

$v=\sqrt228 = 15.10 m/s$

The speed of the ball = 15.10m/s

8 0

2 years ago

A student holds one end of a thread, which is wrapped around a cylindrical spool, as shown above. The student then drops the spo

lesya [120]

by energy conservation we know that

KE or rotation + KE of translation = gravitational PE

now we have

$\frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 = mgH$

also we know that

$v = R\omega$

now we have

$\frac{1}{2}(\frac{1}{2}mR^2)\omega^2 + \frac{1}{2}m(R\omega)^2 = mgH$

$\frac{3}{4}mR^2\omega^2 = mgH$

$\omega = \sqrt{\frac{4gH}{3}}/R$

now when it is rolling on ground the torque acting on it due to friction force is given by

$\tau = R F_f$

$\tau = \mu mg R$

$\alpha = \frac{\mu mg R}{\frac{1}{2}mR^2}$

$\alpha = \frac{2 \mu g}{R}$

now angular speed at any time is given as

$\omega = \omega_i + \alpha t$

$\omega = \sqrt{\frac{4gH}{3}}/R -\frac{2 \mu g}{R} t$

so above is the angular speed in terms of time "t"

7 0

4 years ago

Bohr believed that light is emitted by an electron when...

nignag [31]

The electron fell down to the lowest orbital it could inhabit

4 0

3 years ago

Predict how the wavelength of waves traveling with the same speed would change if the frequency of the waves increases?

postnew [5]

The wavelength would decrease because frequency is inversly proportional to the wavelength provided the speed is constant: f=v/λ

6 0

3 years ago

9 The diagram shows a uniform beam PQ. The length of the beam is 3.0 m and its weight is 50 N. The beam is supported on a pivot

tankabanditka [31]

equilibrium = 1/1

50 N/x = 1/1

x = 1/1 × 50 N

x = 50 N (B)

#LearnWithEXO

6 0

3 years ago