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3 years ago

A toy car on a string is pulled across a table horizontally. The string is at an angle

Physics

1 answer:

ddd [48]3 years ago

6 0

Which way is it being pulled?

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5. Graph A plots a race car's speed for 5 seconds. The car's

Nastasia [14]

The answer for the cars speed is tue

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3 years ago

With the aid of a string, a gyroscope is accelerated from rest to 16 rad/s in 0.40 s. what is its angular acceleration in rad/s2

vovikov84 [41]

Angular acceleration = (change in angular speed) / (time for the change)

Change in angular speed = (ending speed) minus (starting speed)

Change in angular speed = (16 rad/s) - (zero) = 16 rad/s .

Angular acceleration = (16 rad/s) / (0.4 s)

(Average) angular acceleration = 40 rad/s²

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3 years ago

What is true about the force acting on the cars at the end of a roller coaster ride as they slow down right before stopping?

rjkz [21]

That the gravity dosent Change

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3 years ago

Read 2 more answers

A 50-cm-long spring is suspended from the ceiling. A 330 g mass is connected to the end and held at rest with the spring unstret

Nataly [62]

Answer:

a)32.34 N/m

b)10cm

c)1.6 Hz

Explanation:

Let 'k' represent spring constant

'm' mass of the object= 330g =>0.33kg

a) in order to find spring constant 'k', we apply Newton's second law to the equilibrium position 10cm below the release point.

ΣF=kx-mg=0

k=mg / x

k= (0.33 x 9.8)/ 0.1

k= 32.34 N/m

b) The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object.

Therefore, amplitude of the oscillation is 10cm

c)frequency of the oscillation can be determined by,

f= 1/2π $\sqrt{\frac{k}{m} }$

f= 1/2π $\sqrt{\frac{32.34}{0.33} }$

f= 1.57

f≈ 1.6 Hz

Therefore, the frequency of the oscillation is 1.6 Hz

5 0

3 years ago

A student of weight 652 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude

miskamm [114]

Answer:

Explanation:

Given

Weight of person $W=652\ N$

At highest point Magnitude of the normal force $F=585\ N$

net force at highest point

$F_{net}=W+F_c$

where $F_c=$ centripetal force

$F_c=\dfrac{mv^2}{r}$

$F_{net}=F_{n}=$ Normal Force

$585=652+F_c$

$F_c=-67\ N$

Negative sign shows force is in upward direction

At bottom point centripetal force is towards the bottom

$F_n=F_c+W$

$F_n=652+67$

$F_n=719\ N$

8 0

4 years ago