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ziro4ka [17]
3 years ago
5

An object moves from the position +16m to the position +43m in 12s. What us the total displacement

Physics
1 answer:
NeX [460]3 years ago
3 0

First method

initial distance = 16m

final distance= 43 m

total distance covered= final -initial

                                     =43m -16m

                                     =27m

Second method

Si= 16m

Sf =43 m

t= 12 s

first we will find V

V =  (Sf-Si)/ t

V =( 43- 16)/ 12

V = 27/12  ⇒ V= 9/4

V= distance / time

distance= V×time

distance = (9/4) ×12

distance =27

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Answer:

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E xplanation:

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so the value of remainder of frequency f2 and fo is equal to 1:3.

⇒\frac{f2} {f1} = \frac sqrt{m1}[m2}

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This graph indicates that the kinetic energy of an object is increasing. What is the most reasonable explanation for this observ
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B) The object's velocity doubled.

Explanation:

The graph is missing: find it in attachment.

The kinetic energy of an object is the energy possessed by the object due to its motion. It is calculated as

K=\frac{1}{2}mv^2

where

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v is the velocity of the object

We notice that:

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In the graph, one of the two quantities (either mass or speed) is represented on the x-axis, while the quantity on the y-axis is the kinetic energy.

First of all, we notice that the relationship is not linear: this means that the quantity on the x-axis cannot be the mass, so it must be the velocity.

Moreover, we notice that when the quantity on the x-axis increases from 1 to 2 (so, it doubles), the kinetic energy increases by a factor of 4. This means that the object's velocity has doubled, therefore

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3 years ago
The blades in a blender rotate at a rate of 6100 rpm. When the motor is turned off during operation, the blades slow to rest in
MissTica

Answer:

<em>155.80rad/s</em>

Explanation:

Using the equation of motion to find the angular acceleration:

\omega_f = \omega_i + \alpha t

\omega_f is the final angular velocity in rad/s

\omega_i  is the initial angular velocity in rad/s

\alpha is the angular acceleration

t is the time taken

Given the following

\omega_f = 6100rpm

Time = 4.1secs

Convert the angular velocity to rad/s

1rpm = 0.10472rad/s

6100rpm = x

x = 6100 * 0.10472

x  = 638.792rad/s

\omega_f = 638.792rad/s\\

Get the angular acceleration:

Recall that:

\omega_f = \omega_i + \alpha t

638.792 = 0 + ∝(4.1)

4.1∝ = 638.792

∝ = 638.792/4.1

∝ = 155.80rad/s

<em>Hence the angular acceleration as the blades slow down is 155.80rad/s</em>

5 0
2 years ago
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