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3 years ago

If chris throws a baseball 60 meters in 4 seconds, what is the average speed of the football?

Physics

1 answer:

IceJOKER [234]3 years ago

6 0

The average speed of the football is 15 meters per second. Just divide both of the numbers by 4 :)

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Otion

Scorpion4ik [409]

Answer:

SKID

Explanation:

In general, airplane tracks are flat, they do not have cant, consequently the friction force is what keeps the bicycle in the circle.

Let's use Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis.

Y axis y

N- W = 0

N = W

X axis (radial)

fr = m a

the acceleration in the curve is centripetal

a = $\frac{v^2}{r}$

the friction force has the expression

fr = μ N

we substitute

μ mg = m v²/r

v = $\sqrt{\mu g r}$

we calculate

v = $\sqrt{0.1 \ 9.8 \ 3}$

v = 1,715 m / s

to compare with the cyclist's speed let's reduce to the SI system

v₀ = 18 km / h (1000 m / 1 km) (1 h / 3600 s) = 5 m / s

We can see that the speed that the cyclist is carrying is greater than the speed that the curve can take, therefore the cyclist will SKID

5 0

3 years ago

A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result

vivado [14]

Answer:

$21.6\ \text{kg m}^2$

$3.672\ \text{Nm}$

$54.66\ \text{revolutions}$

Explanation:

$\tau$ = Torque = 36.5 Nm

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2$

Torque is given by

$\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2$

The wheel's moment of inertia is $21.6\ \text{kg m}^2$

t = 60.6 s

$\omega_i$ = 10.3 rad/s

$\omega_f$ = 0

$\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2$

Frictional torque is given by

$\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}$

The magnitude of the torque caused by friction is $3.672\ \text{Nm}$

Speeding up

$\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}$

Slowing down

$\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}$

Total number of revolutions

$\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}$

$\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}$

The total number of revolutions the wheel goes through is $54.66\ \text{revolutions}$ .

3 0

3 years ago

Sound wave A has a lower frequency than sound wave B, but both waves

Keith_Richards [23]

Answer: C

Explanation: Amplitude controls loudness, and frequency controls pitch. The more frequent the higher pitch.

4 0

3 years ago

Read 2 more answers

Plsssss I need te answer quick

jok3333 [9.3K]

Explanation:

reflection ... . .......

3 0

3 years ago

Read 2 more answers

A circular coil of wire having a diameter of 20.0 cm and 3000 turns is placed in the earth's magnetic field with the normal of t

Bess [88]

Explanation :

It is given that,

Diameter of the coil, d = 20 cm = 0.2 m

Radius of the coil, r = 0.1 m

Number of turns, N = 3000

Induced EMF, $\epsilon=1.5\ V$

Magnitude of Earth's field, $B=10^{-4}\ T$

We need to find the angular frequency with which it is rotated. The induced emf due to rotation is given by :

$\epsilon=NBA\omega$

$\omega=\dfrac{\epsilon}{NBA}$

$\omega=\dfrac{1.5}{3000\times 10^{-4}\times \pi (0.1)^2}$

$\omega=159.15\ rad/s$

So, the angular frequency with which the loop is rotated is 159.15 rad/s. Hence, this is the required solution.

3 0

3 years ago