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Anastaziya [24]

2 years ago

15

A basketball is thrown for a free throw at 10m/s an angle of 20 degrees, makes it into the hoop and lands on the ground. Find th

e initial velocity in the y direction. ??????????

2 answers:

Kruka [31]2 years ago

6 0

Answer:

2

Explanation:

because uWU

SVETLANKA909090 [29]2 years ago

3 0

Answer:no answer

Explanation:

Cause ur weird

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A resistor has four colored stripes in the following order: orange, orange, brown and silver. What is the resistance of the resi

zubka84 [21]

Answer:

Resistance =330 Ω

Tolerance = 33 Ω

Explanation:

see attached resistor color code table

The first stripe is orange, which means the leftmost digit is a 3.

The second stripe is orange , which means the next digit is a 3.

The third stripe is brown. Since brown is 1, it means add one zero to the right of the first two digits.

The resistance is:

orange-orange-brown= 330 Ω

The tolerance is:

The fourth color band indicates the resistor's tolerance. Tolerance is the percentage of error in the resistor's resistance.

silver is 10%

A 330 Ω resistor has a silver tolerance band.

<em>Tolerance = value of resistor x value of tolerance band </em>

= 330 Ω x 10% = 33 Ω

330 Ω stated resistance +/- 33 Ω tolerance means that the resistor could range in actual value from as much as 363 Ω to as little as 297 Ω.

7 0

3 years ago

Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of

kondaur [170]

Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

$m=\dfrac{F}{g}$

$m=\dfrac{160\ lb}{32\ ft/s^2}$

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

$m=\dfrac{F}{g}$

$m=\dfrac{1.9\ lb}{32\ ft/s^2}$

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

$W=2300\ kg\times 32\ ft/s^2=73600\ lb$

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

$W=0.022\ kg\times 32\ ft/s^2=0.704\ lb$

Hence, this is the required solution.

5 0

3 years ago

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

7 0

3 years ago

Pairs of magnets are shown in the diagram.

belka [17]

The answer is X. I hope this helped

8 0

3 years ago

Read 2 more answers

A rock is dropped from a garage roof from rest. the roof is 6.0 m from the ground. determine the velocity of the rock as it hits

Dmitrij [34]

From rest, a rock is dropped from a garage roof. The roof is 6.0 meters above ground level. The rock will reach the earth at a speed of 10.849 meters per second.

<h3>What is velocity?</h3>

The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity.

it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.

Given data:

V(Final velocity)=? (m/sec)

h(height)= 6.0 m

u(Initial velocity)=0 m/sec

g(gravitational acceleration)=9.81 m/s²

Newton's third equation of motion:

$\rm v_y^2 = u_y^2+ 2gh \\\\\rm v_y^2 = 0+ 2gh\\\\\ v_y= \sqrt{2\times 9.81 \ (m/s^2)\times 6.0 (m)} \\\\ v_y=10.849 \ m/sec$

Hence, the velocity of the rock as it hits the ground will be 10.849 m/sec.

To learn more about the velocity refer to the link ;

brainly.com/question/862972

#SPJ1

7 0

2 years ago