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3 years ago

PLEASE HELP WILL GIVE BRAINLIEST AND EXTRA POINTS!! Super simple I’ve just been sick so I have no clue what’s going on.

Physics

1 answer:

Tju [1.3M]3 years ago

4 0

Answer: 12) 1.07 m/s (right) 13) 4.05 m/s 14) 73 m/s 15) 10.9 m/s

Explanation:

12) Conservation of momentum. Momentum is the produce of mass and velocity.

13(2) + 15(-5) = 13(-5) + 15v

v = 1.06666... ≈ 1.07 m/s (right)

13) 18(9) + 22(0) = 18v + 22v

v = 18(9)/40 = 4.05 m/s

14) 0.65(35) + 0.08(0) = 0.65(26) + 0.08v

v = 73.125

15) This is a bit trickier. Let's ASSUME you jump off at 7 m/s relative to the truck. Doing this, we can assume that the reference frame is moving along with the truck at 10 m/s

the conservation of momentum equation becomes

600(0) + 80(0) = 600v + 80(-7)

v = 0.9333333... m/s

adding back the velocity of the reference frame means the truck is now traveling.

10.9333333... ≈ 10.9 m/s

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Find the equivalent resistance, the current supplied by the battery and the current through each resistor when the specified res

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Answer:

$R_{eq}$ = 19 Ω, I = 0.105 A, V1 = 1.05 V and V2 = 0.95 V

Explanation:

The correct way to solve this type of problem is to find the current or voltage values for the equivalent resistance and from here find the other values.

For a series circuit the equivalent resistance is the sum of the resistance

$R_{eq}$ = R1 + R2

$R_{eq}$ = 10 +9

$R_{eq}$ = 19 Ω

Let's use the equation for the voltage

V = I $R_{eq}$

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3 years ago

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3 years ago

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Read 2 more answers

A major league baseball pitcher throws a pitch that follows these parametric equations:

Alex Ar [27]

Answer:

d) v = 100.2 mph, e) t = 1.25 s, f) -0.0340

Explanation:

d) This is an exercise that we can solve using projectile launch equations, let's start by calculating the time the ball will take to take home-plate

.x = 147 t

t = x / 147

t = 60.5 / 147

t = 0.41156 s

Let's use Pythagoras' theorem to find the speed

v = √ vₓ² + $v_{y}$ ²

vₓ = dx / dt

vₓ = 147

v_{y} = dy / dt

v_{y} = 4 -16 t

We look for speed for the time of arriving at home

$v_{y}$ = 4 - 16 0.41156

v_{y} = -2,585 ft / s

v_{y} = -2.585 ft/ s ( 1 mile /5280 foot) (3600s/1h)

Let's calculate the speed

v = √ (147² + 2,585²)

v = 147.02 ft / s

v = 147.0 ft/s (1 mile/5280 feet)(3600s/1h)

v = 100.2 mph

e) the time it takes for the ball to reach the floor and = 0 foot

y = 5 + 4 t - 16 t²

0 = 5 + 4t - 16t²

t² –t / 4 -5/4 = 0

t² -0.25 t -1.25 = 0

We solve the equation and second degree

t = [0.25 ±√(0.25² + 4 1.25)] / 2

t = [0.25 ± 2.25] / 2

t₁ = 1.25 s

t₂ = -1 s

The positive time is correct

t = 1.25 s

f) The angle of speed when the ball passes home

tan θ = $v_{y}$ / vₓ

θ = tan⁺¹ (v_{y} / vx)

The distance x is given in the exercise

x = 60.5 foot

vₓ = 147 foot / s

The speed y is t = 1.25 s

v_{y} = 5 + 4 1.25 - 16 1.25²

v_{y} = -15 foot / s

θ = tan⁻¹ (-15/147)

θ = -1,947º = -0,0340 rad

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