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3 years ago

Which cars have the same magnitude of momentum? Check all that apply. car 1 car 2 car 3 car 4 car 5

Physics

1 answer:

wariber [46]3 years ago

7 0

Answer:

Car 1 and Car 2 have the same momentum!

Explanation:

Using the formula of momentum (P=m*v), we get for each car:

Car 1: 5kg*2.2m/s = 11kg*m/s

Car 2: 5.5kg*2m/s = 11kg*m/s

Car 3: 6kg*1.35m/s = 8.1kg*m/s

Car 4: 6.5kg*1.9m/s = 12.35kg*m/s

Car 5: 7kg*1.25m/s = 8.75kg*m/s

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If you are doing a "rowing" motion, what muscle of the shoulder are you using?

Andrej [43]

The upper back muscles being worked while using a rowing machine .your upper trapezius and rhomboids located between your shoulder blades, and latissimus dorsi located beneath the armpits

6 0

3 years ago

A farmhand pushes a 26-kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 88 N on the hay, how

Tamiku [17]

Answer:

W = 343.2 J

Explanation:

Given that,

Mass of bale of hay = 26 kg

Horizontal force exerted = 88 N

Distance moved, d = 3.9 m

Work done, W = Fd

Put all the values,

W = 88 N × 3.9 m

= 343.2 J

So, the work done is 343.2 J.

7 0

3 years ago

Consider the circuit shown in the figure to find the power delivered to 6 Ohm resistance (in W). Given that Vs= 30

vekshin1

66.7 Watts

Explanation:

Let $R_{1}=1.0$ ohms, $R_{2}=3.0$ ohms and $R_{3}=6.0\:$ ohms. Since $R_{2}$ and $R_{3}$ are in parallel, their combined resistance $R_{23}$ is given by

$\dfrac{1}{R_{23}}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}$

$R_{23}=\dfrac{R_{2}R_{3}}{R_{2}+ R_{3}}=2.0\:ohms$

The total current flowing through the circuit I is given

$I=\dfrac{V_{s}}{R_{Total}}$

where

$R_{Total}=R_{1}+R_{23}= 3.0\:ohms$

Therefore, the total current through the circuit is

$I=\dfrac{30\:V}{3.0\:ohms}=10\:A$

In order to find the voltage drop across the 6-ohm resistor, we first need to find the voltage drop across the 1-ohm resistor $V_{1}$ :

$V_{1}=(10\:A)(1.0\:ohms)=10\:V$

This means that voltage drop across the 6-ohm resistor $V_{3}$ is 20 V. The power dissipated P by the 6-ohm resitor is given by

$P=I^{2}R_{3}= \dfrac{V^{2}}{R_{3}}=\dfrac{(20\:V)^{2}}{6\:ohms}= 66.7 W$

4 0

3 years ago

An individual is nearsighted; his near point is 13.0 cm and his far point is 50.4 cm. what lens power is needed to correct his n

Colt1911 [192]

Answer:

= -1.984 Diopters

Explanation:

The lens should form an upright, virtual image at the far point.

Therefore;

The image distance, V will be -50.4 cm, since the image is virtual.

For objects at very far distant the object distance,u, will be ∞ ; This means that focal length, f, will be equivalent to image distance, v, that is -50.4 cm

Therefore; f = -50.4/100 = -0.504 m

But, since Power of a lens, P, is given by the reciprocal of focal length in meters, (1/f)

Then, power will be given by;

Power = 1/f

= 1/-0.504 m

=- 1.984

Power is measured in Diopters

Hence = -1.984 Diopters

5 0

3 years ago

Bird bones have air pockets in them to reduce their weight–this also gives them an average density significantly less than that

lakkis [162]

Answer:

a)39ml

b)39g

c)1.1g/ml

Explanation:

Hello!

To solve this exercise use the following steps

1. When Archimedes discovered how to determine the irregular volume of an object by weighing it in the air and in an algua, he found that its volume is equal to the ratio between the differences of the masses (heavy in the air and in the water) and the density of the water (= 1g / ml)

$V=\frac{43g-3.6g}{1g/ml} =39.4ml$

2.as the principle of archimedes says, the displaced volume of water is equal to the volume of the bone which means that 39.4ml of water was displaced, taking into account that the density of water is the ratio between mass and volume we can determine the displaced body of water

$density=\frac{m}{V} \\m=V(density)\\m=(39ml)(1g/ml)=39g$

3. we use the density equation to find the bone density

$density=\frac{43g}{39ml} =1.1g/ml$

8 0

3 years ago