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3 years ago

15

a car travelling at 50m/h on a horizontal highway (a) if the coefficient of static friction between road and tyres on a rainy da

y is 0.100, what is the minimum distance in which the car will stop?

1 answer:

aleksandrvk [35]3 years ago

3 0

Hope this helps!

-Lilly

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Is b= acid + base - water + salt.

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Since rods are about 1000 times more sensitive than cones (at 470 nm), they should be able to detect smaller values of the elect

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a. Rods are about 1000 times more sensitive than cones.

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The answer choice shown is a direct copy of the first line of the problem statement. It doesn't appear to be any more complicated than that. (It's a reading comprehension question.)

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After oxygen has been administered, the next priority intervention the nurse would initiate for a patient with a pulmonary embol

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After oxygen has been administered, the next priority intervention the nurse would initiate for a patient with a pulmonary embolus is the administration of IV Heparin.

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2 years ago

You are making a telephone out of two aluminum cans and some string. You can choose between two types of string: a 2-m length of

Nataliya [291]

Answer:

C)You should use the thin cooking twine.

Explanation:

A)You can choose either because they are the same length and will produce the same wave speed.

B)You should use the heavy rope.

C)You should use the thin cooking twine.

The speed of wave in a string is given by the following formula:

| $v$ | = $\sqrt{\frac{F_T}{u} }$

Where | $v$ | = speed of wave, $F_T$ = tension in the string, and μ = mass per length of the string.

<em>Even though the two strings have the same length, the μ (mass/length) for the heavy rope will be more than the that of a thin rope. Consequently, the </em> $F_T$ <em>:μ for the thin rope will be higher than that of the heavy rope and as such, gives a bigger |</em> $v$ <em>|. </em>

Therefore, the thin rope should be used in order to get a faster wave speed in the telephone.

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3 years ago

A 250 kg flatcar 25 m long is moving with a speed of 3.0 m/s along horizontal frictionless rails. A 61 kg worker starts walking

Anna11 [10]

Answer:

x=31.09m

Explanation:

p1=p2

The momentum of flatcar and the momentum of the worker so

The velocity of the worker is:

$m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}$

The total motion has a total velocity and is

$Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}$

The time the worker take walking is

$t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s$

Now the total time and the total velocity determinate the motion of tha flatcar how far has moved

$x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m$

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3 years ago