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dybincka [34]
3 years ago
11

The impulse experienced by a body is equivalent to the body’s change in?

Physics
1 answer:
vovangra [49]3 years ago
5 0
<span>impulse =force*time=mass*acceleration*time=mass*... in momentum , I hope this helps you out!! Also have an amazing day and good luck on any further work !!!

        #brainlyzkool</span>
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Water is most dense at 4 degrees Celsius. Since at this temperature 1 ml of water has a mass of 1 g. What is its density?
sveticcg [70]
The density is 1. 1/1=1
7 0
3 years ago
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A student has the following equipment - copper wire, nail made of iron,a battery, paperclips.
ahrayia [7]
The first step would be to create an electromagnet. You can create an electromagnet by winding a copper wire around the nail, the connect both ends to the battery. A current would start flowing around the nail through the wire, creating an electromagnet with its own magnetic field. Next, bringing the electromagnet to the mixture of copper and iron would slowly attract the pieces of iron (as copper is weakly magnetic). Do this slowly and the iron pieces would all slowly be separated from the copper pieces.
4 0
3 years ago
A truck has shock absorbers with a spring constant of 24200 N/m. When it hits a bump, it oscillates at 0.429 Hz. What is the mas
siniylev [52]

Answer:

3331.5 kg

Explanation:

Given:

Spring constant of the spring (k) = 24200 N/m

Frequency of oscillation (f) = 0.429 Hz

Let the mass be 'm' kg.

Now, we know that, a spring-mass system undergoes Simple Harmonic Motion (SHM). The frequency of oscillation of SHM is given as:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

Rewrite the above equation in terms of 'm'. This gives,

2\pi f=\sqrt{\frac{k}{m}}\\\\Squaring\ both\ sides,\ we\ get:\\\\(2\pi f)^2=\frac{k}{m}\\\\m=\frac{k}{4\pi^2 f^2}

Now, plug in the given values and solve for 'm'. This gives,

m=\frac{24200\ N/m}{4\pi^2\times (0.429\ Hz)^2 }\\\\m=\frac{24200\ N/m}{4\pi^2\times 0.184\ Hz^2}\\\\m\approx3331.5\ kg

Therefore, the mass of the truck is 3331.5 kg.

3 0
3 years ago
A car travels 15 kilometers west in 10 minutes. After reaching the destination, the car travels back to the starting point, agai
jeka94

Speed = (distance traveled) / (time to travel the distance).
 
Strange as it may seem, 'velocity' is completely different. 

Velocity doesn't involve the total distance traveled at all. 
Instead, 'velocity' is based on 'displacement' ... the distance
between the start-point and end-point, regardless of the route
taken to get there.  So the displacement in driving once around
any closed path is zero, because you end up where you started. 

Velocity =

           (displacement during some time)
divided by
            (time for the displacement)

AND the direction from the start-point to the end-point.


For the guy who drove 15 km to his destination in 10 min, and then
back to his starting point in 5 min, (assuming he returned by way of
the same 15-km route):

         Speed = (15km + 15km) / (10min + 5min)  =  (30/15) (km/min)

                                                                                 =  2 km/min.

        Velocity = (end location - start position) / (15 min)  =  Zero .

5 0
3 years ago
What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?
IgorLugansk [536]

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as m_{1} \ and\ m_{2}\ respectively

Let\ m_{1} =1 kg\ and\ m_{2} =2 kg

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

                             m_{1} g-T=m_{1} a        [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction [\mu]=0.13

Hence\ the\ frictional\ force\ F=\mu N

                                                     F=\mu m_{2} g

Hence the net force acting on the body having mass 2 kg-

                                  T-\mu m_{2} g=m_{2} a  [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

                           m_{1} g-T=m_{1}a             [1]

                           T-\mu m_{2}g= m_{2} a   [2]

                           ---------------------------------------------------

                           [m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a

                           a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g

                           a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2

                           a=\frac{0.74}{3} *9.8\ m/s^2

                           a=2.417 m/s         [ans]

6 0
3 years ago
Read 2 more answers
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