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3 years ago

Which of these are part of our solar system? select all that apply

Physics

2 answers:

Len [333]3 years ago

6 0

A)planets
b)the sun
c)moons
e)comets
f)asteroids

Papessa [141]3 years ago

4 0

Everything on this list is part of the solar system EXCEPT Polaris.

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Today's topic

babunello [35]

Answer:

While a body is said to be in motion if it changes its position with respect to immediate surroundings.

A body is said to be in uniform motion if it covers equal distances in equal interval of time.

A body is said to be in non-uniform motion if it covers unequal distances in equal interval of time or vice-versa

4 0

2 years ago

Which of the following would be the magnitude of the vector given a horizontal component of 30 and a

Scorpion4ik [409]

Answer:

Explanation:

Use the Pythagorean theorem to find the length of the diagonal, or the hypotenuse of an imaginary triangle. 30^2 + 40^2 = 2500, which is 50^2. So, the magnitude is 50.

Brainliest, please :)

7 0

1 year ago

If an object travels at a constant speed in a circular path, the acceleration of the object is:

Vilka [71]

Answer:

1- The acceleration of the object is larger in magnitude the smaller the radius of the circle.

Explanation:

The acceleration of an object in a circular path is

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

As can be seen from the equation, if the radius of the circle is decreases, the magnitude of the acceleration increases.

As for the direction of the acceleration, it is always towards the center, and it is always perpendicular to the direction of the velocity.

6 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

2 years ago

Is there supposed to be a slit at the top of your adams apple?

Allushta [10]

No, there isn't. Please consult your doctor if this is the case with yours or someone you know.

7 0

2 years ago