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Oksi-84 [34.3K]
2 years ago
12

A softball pitcher throws a softball to a catcher behind home plate. the softball is 3 feet above the ground when it leaves the

pitcher’s hand at a velocity of 50 feet per second. if the softball’s acceleration is –16 ft/s2, which quadratic equation models the situation correctly? h(t) = at2 vt h0 h(t) = 50t2 – 16t 3 h(t) = –16t2 50t 3 3 = –16t2 50t h0 3 = 50t2 – 16t h0
Physics
1 answer:
Bumek [7]2 years ago
6 0

The correct option is (B) h(t) = -16t2 + 50t + 3

An equation like this has the generic form h(t) = at2 + v0t + h0,

where; a is the gravitational constant,

v0 is the initial velocity, and

h0 is the initial height.

Given that the gravitational constant is -16.

The equation h(t) = -16t2 + 50t + 3 is given by the starting velocity of 50 and the initial height of 3.

Learn more about Projectile with the help of the given link:

brainly.com/question/11422992

#SPJ4

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A ball rolls horizontally off a table and a height of 1.4 m with a speed of 4 m/s. How long does it take the ball to reach the g
Hitman42 [59]

For vertical motion, use the following kinematics equation:

H(t) = X + Vt + 0.5At²

H(t) is the height of the ball at any point in time t for t ≥ 0s

X is the initial height

V is the initial vertical velocity

A is the constant vertical acceleration

Given values:

X = 1.4m

V = 0m/s (starting from free fall)

A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)

Plug in these values to get H(t):

H(t) = 1.4 + 0t - 4.905t²

H(t) = 1.4 - 4.905t²

We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:

1.4 - 4.905t² = 0

4.905t² = 1.4

t² = 0.2854

t = ±0.5342s

Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))

t = 0.53s

8 0
3 years ago
Read 2 more answers
Calculate the density of a tin of mass 100g whose dimensions are 2cmx5cmx​
Luda [366]

Answer:

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7 0
3 years ago
A charge of -8.5 µC is traveling at a speed of 9.0 106 m/s in a region of space where there is a magnetic field. The angle betwe
Sindrei [870]

Answer:

The magnitude of the magnetic field is 9.3\times 10^{-5}\ T.

Explanation:

Given that,

Charge, q=-8.5\ \mu C=-8.5\times 10^{-6}\ C

Speed of the charged particle, v=9\times 10^6\ m/s

The angle between the velocity of the charge and the field is 56°.

The magnitude of force, F=5.9\times 10^{-3}\ N

We need to find the magnitude of the magnetic field. When a charged particle moves in the magnetic field, the magnetic force is experienced by it. The force is given by :

F=qvB\ \sin\theta

B is the magnetic field.

B=\dfrac{F}{qv\ \sin\theta}\\\\B=\dfrac{5.9\times 10^{-3}}{8.5\times 10^{-6}\times 9\times 10^6\ \sin(56)}\\\\B=9.3\times 10^{-5}\ T

So, the magnitude of the magnetic field is 9.3\times 10^{-5}\ T. Hence, this is the required solution.

4 0
3 years ago
Name and explain two sleeping disorders
vfiekz [6]
Insomnia and night terrors
4 0
3 years ago
What is the heat capacity of an object at 25.5∘C that absorbs 45 kJ of heat and is heated to 28.2∘C?
sergey [27]

Answer:

16.6 kJ/°C

Explanation:

given,

Amount of heat absorbed = 45 kJ

initial temperature, T₁ = 25.5°C

final temperature, T₂ = 28.2°C

change in temperature = T₂ - T₁

                                       = 28.2 - 25.5  = 2.7° C

Heat\ capacity = \dfrac{Heat\ absorbed}{\Delta T}

Heat\ capacity = \dfrac{45\ kJ}{2.7}

Heat\ capacity = 16.6\ kJ/^0C

Heat capacity of the object is equal to 16.6 kJ/°C

4 0
3 years ago
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