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3 years ago

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A marble at the front of a truck bed traveling at 20 m/s relative to the highway rolls toward the back of the truck bed with a s

peed of 7m/s relative to the truck bed. What is the velocity of the marble relative to the highway?

1 answer:

irga5000 [103]3 years ago

5 0

14 m/s in the direction of the truck

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A massless string connects a 10.00 kg mass to a 13.00 kg cart which is resting on a frictionless horizontal surface. The mass ha

ch4aika [34]

The cart's acceleration to the right after the mass is released is determined as 7.54 m/s².

<h3>Acceleration of the cart</h3>

The acceleration of the cart is determined from the net force acting on the mass-cart system.

Upward force = Downward force

ma = mg

13a = 10(9.8)

13a = 98

a = 98/13

a = 7.54 m/s²

Thus, the cart's acceleration to the right after the mass is released is determined as 7.54 m/s².

Learn more about acceleration here: brainly.com/question/14344386

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2 years ago

Are cell phones scientific

Aleonysh [2.5K]

Yes, because cell phones use scientific structure to build.

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3 years ago

Read 2 more answers

A spring with spring constant k is suspended vertically from a support and a mass m is attached. The mass is held at the point w

garik1379 [7]

Answer:

The oscillation frequency of the spring is 1.66 Hz.

Explanation:

Let m is the mass of the object that is suspended vertically from a support. The potential energy stored in the spring is given by :

$E_s=\dfrac{1}{2}kx^2$

k is the spring constant

x is the distance to the lowest point form the initial position.

When the object reaches the highest point, the stored potential energy stored in the spring gets converted to the potential energy.

$E_P=mgx$

Equating these two energies,

$\dfrac{1}{2}kx^2=mgx$

$\dfrac{k}{m}=\dfrac{2g}{x}$ .............(1)

The expression for the oscillation frequency is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{2g}{x}}$ (from equation (1))

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{2\times 9.8}{0.18}}$

f = 1.66 Hz

So, the oscillation frequency of the spring is 1.66 Hz. Hence, this is the required solution.

8 0

4 years ago

Polaczono szeregowo cewke ul=120 rezystor ur=100 i kondensator uc=50 Napiecie miezone voltomierzem nie moze byc: u>100 u<1

romanna [79]

Answer:

<u>Translated from Polish language:</u>

Coil connected in series ul = 120 resistor ur = 100 and capacitor uc = 50 Voltage with voltmeter must not be:

The voltmeter must not be connected in series rather it should be connected in parallel to the circuit to provide much effective results for analyzing the circuits potential difference.

3 0

3 years ago

Rank the angular speed of the following objects, from highest to lowest:- A bowling ball of radius 12.3 cm rotating at 8.21 radi

dlinn [17]

Answer:

A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s)

Explanation:

<u>To rank the angular speed (ω) of the objects, we need first calculate its value for every object:</u>

A bowling ball of radius 12.3cm rotating at 8.21 radians per second:

ω = 8.21 rad/s

A tire of radius 0.321m rotating at 75.8 rpm:

$\omega = 75.8 \frac{rev}{min}\cdot \frac{2\pi rad}{1rev}\cdot \frac{1min}{60s} = 7.94rad/s$

A 6.84cm diameter top spinning at 375 degrees per second:

$\omega = 375 \frac{^\circ}{s}\cdot \frac{2\pi rad}{360^ \circ} = 6.54rad/s$

A square with sides (b) 0.458m long, whose corners are moving with tangential speed (v) 2.51 m/s as it rotates about its center:

$\omega = \frac{v}{r} = \frac{v}{\frac{b}{2}\sqrt 2} = \frac{2.51 m/s}{\frac{0.458 m}{2} \sqrt 2} = 7.75rad/s$

A rock on a string, being swung in a circle of radius 0.521 m with a centripetal acceleration (a) of 25.4 m/s²:

$\omega = \sqrt \frac{a}{r} = \sqrt \frac{25.4 m/s^{2}}{0.521m} = 6.98rad/s$

<u>Now, the rank of the angular speed of the objects, from highest to lowest is: </u>

A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s)

I hope it helps you!

4 0

3 years ago