Ask question

Ask question

All categories

3 years ago

11

What velocity must a car with a mass of 1280 kg have in order to have the same momentum as a 2230?

1 answer:

sveta [45]3 years ago

8 0

Momentum = (mv).
<span>(2110 x 24) = 50,640kg/m/sec. truck momentum. </span>
<span>Velocity required for car of 1330kg to equal = (50,640/1330), = 38m/sec</span>

You might be interested in

All of the following were barriers to minority participation in early psychology except

Masja [62]

<span>All of the following were barriers to minority participation in early psychology except school desegregation. The correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer that you were looking for and the answer has come to your great help.</span>

4 0

2 years ago

Read 2 more answers

A positron is an elementary particle identical to an electron except that its charge is . An electron and a positron can rotate

denis23 [38]

Explanation:

According to the energy conservation,

$F_{centripetal} = F_{electric}$

$\frac{mv^{2}}{r} = \frac{kq^{2}}{d^{2}}$

$v^{2} = \frac{kq^{2}r}{d^{2}m}$

= $\frac{9 \times 10^{9} N.m^{2}/C^{2} \times 1.6 \times 10^{-19} C \times 0.75 \times 10^{-9} m}{(1.50 \times 10^{-9}m)^{2} \times 9.11 \times 10^{-31} kg}$

= $8.430 \times 10^{10} m^{2}/s^{2}$

v = $\sqrt{8.430 \times 10^{10} m^{2}/s^{2}}$

= $2.903 \times 10^{5} m/s$

Formula for distance from the orbit is as follows.

S = $2 \pi r$

= $2 \times 3.14 \times 0.75 \times 10^{-9} m$

= $4.71 \times 10^{-9} m$

Now, relation between time and distance is as follows.

T = $\frac{S}{v}$

$\frac{1}{f} = \frac{S}{v}$

or, f = $\frac{v}{S}$

= $\frac{2.903 \times 10^{5} m/s}{4.71 \times 10^{-9} m}$

= $6.164 \times 10^{13} Hz$

Thus, we can conclude that the orbital frequency for an electron and a positron that is 1.50 apart is $6.164 \times 10^{13} Hz$ .

7 0

2 years ago

What is the wavelength of the wave pictured above?

love history [14]

Answer:

its counting by 4 the multi 4,8,12,16

3 0

1 year ago

A 2 kg block is on a horizontal surface. A horizontal force is applied by a person to the block to pull it on the surface with a

Agata [3.3K]

Answer:

work done = 500 J

option e is correct

Explanation:

given data

mass = 2 kg

acceleration = 2 m/s

coefficient of kinetic friction = 0.3

block D = 50 m

to find out

work

solution

we know that work done is

work done = f × d ...........1

here f is force and d is block i.e 50 m

so here

force , f - k mg =ma

f = 2 ( 2 + 0.3 ×10 )

f = 10 N

so from equation 1

work done = f × d

work done = 10 × 50

work done = 500 J

option e is correct

3 0

3 years ago

What does a fire burn a hole right in the middle of the chair???

earnstyle [38]

i dont understand wat ur tryna say ma dude but whatever it is its.............PIZZA

4 0

3 years ago