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Sedaia [141]
3 years ago
13

What is the momentum of a 2.0kg ball rolling at 6.0 m/s

Physics
1 answer:
irga5000 [103]3 years ago
4 0
Momentum can be defined by the formula p=m*V (where m is mass and V is velocity) so if we plug in these numbers:

p = 2kg * 6m/s
p = 12 kgm/s
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Consider the expression below. Assume m is an integer. 6m(2m + 18) Enter an expression in the box that uses the variable m and m
Anika [276]
6x2=12m
6x18=108
12m+108
Simplified: m+9 bc 12/12 and 108/12
8 0
4 years ago
A 5 N force is applied to a 3 kg ball to change its velocity from 9 m/s to 3 m/s. What is the impulse on the ball ?
nika2105 [10]
So, first the formula of Impulse is
I = force * time
We have force but no time.
Then, find time.
Next find acceleration,
F = mass * acceleration
5 = 3 * a
1.67 m/s^2
Next find time,
Acceleration = change in velocity / time
Change in velocity is velocity final - velocity initial
1.67 = 3 - 9 / time
Time = 3.6 s (round to 2 s.f.)
Lastly,
Impulse = force * time
Impulse = 5 * 3.6
Impulse is 18 Ns
3 0
3 years ago
Fast and safe heart rate for workouts is called muscular strength? True or false
Vsevolod [243]

Answer:

False

Explanation:

8 0
3 years ago
Read 2 more answers
Explain imprinting in relation to Lorenz's geese. What would most likely have happened if the first thing the goslings saw was a
iragen [17]

If the first thing that the goslings saw was a dog, they would have followed the dog as a mother.

Imprinting refers to the process of training an animal to bond with anything it sees after birth even if it is not its real mother. Lorenz first achieved imprinting in 1935 using geese which followed him as their mother shortly after they were born.

If the geese were exposed to a dog, they could also have seen the dog as their mother and followed it accordingly shortly after birth.

Learn more: brainly.com/question/11401513

7 0
2 years ago
A particle with a charge of +4.20nC is in a uniform electric field E⃗ directed to the left. It is released from rest and moves t
Morgarella [4.7K]

Answer:

(A). The work done is 1.50\times10^{-6}\ J.

(B). The potential of the starting point with respect to the endpoint is 357.14 V.

(C). The magnitude of E is 5952.38 N/C.

Explanation:

Given that,

Charge = 4.20 nC

Distance = 6.00 cm

Kinetic energy K.E=1.50\times10^{-6}\ J

The particle start from rest.

So, the initial kinetic energy i zero.

(A). We need to calculate the work by the electric force

Using formula of work done

W = \Delta K.E

W=K.E_{f}-K.E_{i}

Put the value into the formula

W= 1.50\times10^{-6}-0

W=1.50\times10^{-6}\ J

The work done is 1.50\times10^{-6}\ J.

(B). We need to calculate the potential of the starting point with respect to the endpoint

We know that.

Change in potential energy = change in kinetic energy

\Delta P.E=\Delta K.E

So, U = 1.50\times10^{-6}

Using formula of potential

V=\dfrac{U}{q}

Put the value into the formula

V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}

V=357.14\ V

The potential of the starting point with respect to the endpoint is 357.14 V.

(C). We need to calculate the magnitude of E

Using formula of work done

W=F\times r....(I)

Using formula of force

F=qE

Put the value in the equation (I)

W=qE\times r

E=\dfrac{W}{q\times r}

Put the value into the formula

E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}

E=5952.38\ N/C

The magnitude of E is 5952.38 N/C.

Hence, This is the required solution.

7 0
3 years ago
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