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JulsSmile [24]
3 years ago
10

In a face, you run 3000 meters east in 21 minutes. What is your velocity in km/min?

Physics
1 answer:
Anni [7]3 years ago
7 0

3 \times  \ 21
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A rolling ball has an initial velocity of 1.6 meters per second. if the ball has a constant acceleration of 0.33 meters per seco
lukranit [14]
We have:

Initial velocity (u) = 1.6 m/s
Constant acceleration (a) = 0.33 m/s²
Time (t) = 3.6 sec

There are five constant acceleration equations that would help us to find the velocity:

v=u+at
s=ut+ \frac{1}{2}at^2
s= \frac{1}{2}(u+v)t
v^2=u^2+2as
s=vt- \frac{1}{2}at^2

Since we have u, a, t and we want v
We will use the first formula v=u+at

v=1.6+(0.33)(3.6)
v= 2.788 m/s
7 0
3 years ago
Speed is usually what
kolbaska11 [484]

Answer:

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Explanation:

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3 0
3 years ago
Which experimental result led to the inference that atoms contained electrons?
Otrada [13]

Well most of the particles did pass through and a few were deflected. but i think the answer is A

3 0
3 years ago
Read 2 more answers
An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic
malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

\omega_f = \omega_0 + \alpha t

Where,

\omega_f =Final Angular Velocity

\omega_0 =Initial Angular velocity

\alpha = Angular acceleration

t = time

The relation between the tangential acceleration is given as,

a = \alpha r

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s

t = 11s

Replacing the previous equation with our values we have,

\omega_f = \omega_0 + \alpha t

9.8436 = 6.5973 + \alpha (11)

\alpha = \frac{9.8436- 6.5973}{11}

\alpha = 0.295rad/s^2

The tangential velocity then would be,

a = \alpha r

a = (0.295)(0.2)

a = 0.059m/s^2

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

\omega_f^2=\omega_0^2+2\alpha\theta

Replacing with our values and re-arrange to find \theta,

\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}

\theta = \frac{9.8436^2-6.5973^2}{2*0.295}

\theta = 90.461rad

That is equal in revolution to

\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev

The linear displacement of the system is,

x = \theta*(2\pi*r)

x = 14.397*(2\pi*\frac{0.25}{2})

x = 11.3m

5 0
3 years ago
A paintball’s mass is 0.0032kg. A typical paintball strikes a target moving at 85.3 m/s.
vekshin1

Answer:

A)  If the paintball stops completely the magnitude of the change in the paintball’s momentum is  p=0.273kg*m/s

B) If the paintball bounces off its target and afterward moves in the opposite direction with the same speed, the change in the paintball’s momentum is  p=0.546kg*m/s

C) A paintball bouncing off your skin in the opposite direction with the same speed hurts more than a paintball exploding upon your skin because of the strength exerted is twice than if it explodes.

Explanation:

Hi

A) We use the formula of momentum p=mv, so we have p=0.0032kg*85.3m/s=0.273kg*m/s

B) We use the same formula above, then due we have a change of direction at the same speed, therefore the change in the momentum is the double so

p=2*0.0032kg*85.3m/s=0.546kg*m/s.

C) The average strength of the force an object exerts during impact is determined by the amount the object’s momentum changes. therefore

F=\frac{\Delta p}{\Delta t}, as we don't have any data about the impact time but we know momentum is twice, time does no matter and strength is twice too.

4 0
3 years ago
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