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ss7ja [257]
3 years ago
13

What speed must a 600 kg car have in order to have the same momentum as a 1200 kg truck traveling at a velocity of 10 m/s to the

west? 5 m/s west B 20 mls west 5 m/s east 20 m/s east
A) 5m/s west
B) 20m/s west
C) 5m/s east
D) 20m/s east
​
Physics
1 answer:
yKpoI14uk [10]3 years ago
4 0

Answer:

20 m/s

Explanation:

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Joaquin is at rest at the top of a hill on a skateboard. Four seconds later, he reaches the bottom of the hill at a final veloci
Andre45 [30]

Explanation:

Given: Initial Velocity=0

Final Velocity= 20 m/s

Time= 4 s

Accleration: Unknowns

Use kinematic equation

v   _{f} = v _{i} + at

20 = 0 + at

\frac{20m}{s}  = a(4s)

a =  \frac{5m}{ {s}^{2} }

so a = 5 m/s^2

8 0
3 years ago
Does a resistor resist current or voltage
KIM [24]
The answer is current
4 0
3 years ago
A rectangular tank is filled to a depth of 10m with freshwater and open to air at atmospheric pressure.
charle [14.2K]

Answer:

<em>1.</em> <em>39068.07 N</em>

<em>2. 19534.036 N</em>

Explanation:

depth of water h = 10 m

atmospheric pressure Patm = 101325 Pa

density of water p = 1000 kg/m^3

acceleration due to gravity g = 9.81 m/s^2

pressure due to depth of water = pgh

P = 1000 x 9.81 x 10 = 98100 Pa

total pressure on the bottom of the tank is Patm + p = 101325 + 98100 = 199425 Pa

Left plug has diameter = 50 cm = 0.5 m

radius = 0.5/2 = 0.25 m

height = 1 cm = 0.01 m

<em>height below tank surface = 10 - 0.01 = 9.99</em>

pressure at this depth =  1000 x 9.81 x 9.99 = 98001.9 Pa

<em>total pressure = Patm + P = 101325 + 98001.9 = 199326.9 Pa</em>

surface area of plug = πr^{2} = 3.142 x 0.25^{2} = 0.196 m^{2}

<em>force required to lift left plug = pressure x area</em>

F =  199326.9 x 0.196 = <em>39068.07 N</em>

<em>The right side is a hemisphere with the same diameter, therefore surface area is half of the left plug</em>

A = 0.196/2 = 0.098 m^{2}

force F required to lift right plug =  199326.9  x 0.098 =<em> 19534.036 N</em>

6 0
3 years ago
The radii of the sprocket assemblies and the wheel of the bicycle in the figure are:
Furkat [3]
To solve this task we have to make a proportion, but firstly we have to set up all the main points : so, the distance is  s=r(B), that has its <span>r=radius,B=angle in rad velocity v=ds/dt= w(r)
Do not forget about </span> w = angular speed in rad/s and w1 = 1 revolution/sec = 2Pi (rad/s)
Now we can go to proportion
v1=v2
w1*r1 = w2r2w2 = w1 * r1/r2 = 2w1 = 4Pi (rad/s)
w2 = w3 (which is the   angular velocity of the rear wheel) &#10;
SOLVING FOR A : v3 = w3 * r3 = 4pi * 14 (inch/s) = 14.66 ft/sec
v3 = 14.66 ft/sec(1 mile/5280 ft)( 3600 sec/h)= 9.99 or something about <span>10 mph --- SOLVING FOR B.
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7 0
3 years ago
A 2.00-kg object A is connected with a massless string across a massless, frictionless pulley to a 3.00-kg object B. Object A re
slamgirl [31]

Answer:

  • tension: 19.3 N
  • acceleration: 3.36 m/s^2

Explanation:

<u>Given</u>

  mass A = 2.0 kg

  mass B = 3.0 kg

  θ = 40°

<u>Find</u>

  The tension in the string

  The acceleration of the masses

<u>Solution</u>

Mass A is being pulled down the inclined plane by a force due to gravity of ...

  F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N

Mass B is being pulled downward by gravity with a force of ...

  F = mg = (3 kg)(9.8 m/s^2) = 29.4 N

The tension in the string, T, is such that the net force on each mass results in the same acceleration:

  F/m = a = F/m

  (T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)

  T = (2(29.4) +3(12.5986))/5 = 19.3192 N

__

Then the acceleration of B is ...

  a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2

The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.

3 0
3 years ago
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