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2 years ago

8

The mass of a basketball is three times greater than the mass of a softball . Compare the momentum’s of a softball and a basketb

all if they both are moving at the same velocity

1 answer:

Lemur [1.5K]2 years ago

4 0

They are moving the softball of basketball

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A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest. with what speed does the goalie slide on the frict

madreJ [45]

Answer:

0.076 m/s

Explanation:

Momentum is conserved:

m v = (m + M) V

(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V

V = 0.076 m/s

After catching the puck, the goalie slides at 0.076 m/s.

8 0

2 years ago

49. A vertically hung 0.50-meter- long spring is stretched from its equilibrium position to a length of 1.00 meter by a weight a

Natali5045456 [20]

Answer:

$K=120$

Explanation:

From the question we are told that

Length of spring $l_1=0.5m$

Length of stretched $l_s=1m$

Potential energy of spring $E=15J$

Generally equation for energy stored is mathematically given as

$U=1/2K \triangle x^2$

$K=\frac{2U}{\triangle x^2}$

$K=\frac{2*15}{ 0.5^2}$

Therefore value of the spring constant in N/m? is given as

$K=120$

4 0

1 year ago

Calculate the volume of a liquid with a density of 5.45 g/ml and a mass of 65g

katrin [286]

Density=mass/volume
5.45g/ml=65g/V
V=65g/5.42g/ml
V=11.92ml

5 0

2 years ago

Consider the vector b⃗ with magnitude 4.00 m at an angle 23.5∘ north of east. what is the x component bx of this vector? express

BlackZzzverrR [31]

Decomposing the vector b on the x-axis and the y-axis, we get a rectangle triangle where the two sides are bx (x-axis) and by (y-axis), and b is the hypothenuse.
The component in x, bx, is equal to the product between the hypothenuse and the cosine of the angle between b and the x-axis, which is $23.5 ^{\circ}$ :
$b_x = b \cos (23^{\circ})=(4.00 m)(\cos (23^{\circ}))=3.68 m$

6 0

2 years ago

. A lightbulb with a resistance of 2.9 ohms is operated using a 1.5-volt battery. At what rate is

Kobotan [32]

Answer:

Explanation:

Given that,

A light bulb has a resistance of 2.9ohms

R = 2.9 ohms

And a battery of 1.5V is applied

V = 1.5 V

We want to find the rate of energy transformed

First we need to know what rate of energy is

Rate of energy implies that we want to find power. Power is the rate at which work is done

P = Workdone / time

Then,

In electronic, the power dissipated by a resistor is given as

P = V² / R

P = 1.5² / 2.9

P = 0.7759 W

P ≈ 0.776 W

So, the rate at which electrical energy transformed in the lightbulb is 0.776 Watts

8 0

2 years ago

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