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LiRa [457]
3 years ago
9

A 50 kg woman, riding on a 10 kg cart, is moving east at 5.0 m/s. the woman jumps off the cart and hits the ground running at 7.

0 m/s, eastward, relative to the ground. calculate the velocity of the cart after she jumps off.
Physics
1 answer:
pav-90 [236]3 years ago
4 0
To answer this question, we will use the law of conservation of momentum which states that:
(m1+m2)Vi = m1V1 + m2V2 where:
m1 is the mass of the woman = 50 kg
m2 is the mass of the cart = 10 kg
Vi is the initial velocity (of woman and cart combined) = 5 m/sec
V1 is the final velocity of the woman = 7 m/sec
V2 is the final velocity of the cart that we need to calculate

Substitute with the givens in the above equation to get the final velocity of the cart as follows:
(50+10)(5) = (50)(7) + (10)V2
10V2 = -50
V2 = -5 m/sec
Note that the negative sign indicates that the cart is moving in an opposite direction to the others.
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Number 5 and 6 i need just give answer
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Explanation:

voltage = current × resistance

5.

12 V = 4.2 A × resistance

resistance = 12 V / 4.2 A = 2.857142857... Ohm

FYI :

4.2 A would be a lot for a small electronic device like a CD player. that would be 12×4.2 = 50.4 Watt, and the CD player would get really hot.

6.

120 V = current × 12 Ohm

current = 120 V / 12 Ohm = 10 A

8 0
2 years ago
The dimension of Radius of gyration​
serg [7]
Answer: So finally, the dimensional formula of the radius of gyration will be written as: [M0LT0]. The power of zero on the dimension of the mass and time shows that the mass and the time dimensions are zero for the radius of gyration. Hope this helps (:
4 0
3 years ago
A machinist with normal vision has a near point at 25 cm. The machinist wears eyeglasses in order to do close work. The power of
Alik [6]

Answer:

17.4 cm

Explanation:

Power of lens = +1.75 diopters

Focal length of lens

f=\frac{100}{1.75}=\frac{400}{7}

This is a convex lens as focal the diopter given is positive which makes the focal length positive. Image distance will be negative.

v = -25

\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\Rightarrow \frac{1}{-25}-\frac{1}{u}=\frac{1}{\frac{400}{7}}\\\Rightarrow -\frac{1}{u}=\frac{7}{400}+\frac{1}{25}\\\Rightarrow \frac{1}{u}=-\frac{7}{400}-\frac{1}{25}\\\Rightarrow \frac{1}{u}=-0.0575\\\Rightarrow u=-17.4\ cm

∴ The new near point is 17.4 cm

7 0
3 years ago
If the distance d (in meters) traveled by an object in time t (in seconds) is given by the formula d=a+bt2, the si units of a an
ss7ja [257]
Distance , d = a+bt^2

The unit of d is in meter and t is in seconds.

So the unit of a a must be meter.

Now we have unit of bt^2 is meter.

So unit of b*second^2 = meter

Unit of b = meter/second^2

So unit of a = m and unit of b = m/s^2.
8 0
3 years ago
A 1300 kg car moving at 20 m/s and a 900 kg car moving at 15 m/s in precisely oppositedirections participate in a head-on crash.
miskamm [114]

Given

Car 1

m1 = 1300 kg

v1 = 20 m/s

m2 = 900 kg

v2 = -15 m/s

(Negative sign shows that direction of car 2 is opposite to car 1)

Procedure

As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,

\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{1300\operatorname{kg}\cdot20m/s-900\operatorname{kg}\cdot15m/s}{1300\operatorname{kg}+900\operatorname{kg}} \\ v=5.681m/s \end{gathered}

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.

5 0
1 year ago
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