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3 years ago

I NEED HELP SOLVING THIS!!!!!!!!!!!!

Physics

1 answer:

Neko [114]3 years ago

7 0

Answer:

Yes

Explanation:

The given parameters are;

The speed with which the fastball is hit, u = 49.1 m/s (109.9 mph)

The angle in which the fastball is hit, θ = 22°

The distance of the field = 96 m (315 ft)

The range of the projectile motion of the fastball is given by the following formula

$Range = \dfrac{u^2 \times sin(2\cdot \theta)}{g}$

Where;

g = The acceleration due to gravity = 9.81 m/s², we have;

$Range = \dfrac{49.1^2 \times sin(2\times22^{\circ})}{9.81} \approx 170.71 \ m$

Yes, given that the ball's range is larger than the extent of the field, the batter is able to safely reach home.

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It has been suggested that a heat engine could be developed that made use of the fact that the temperature several hundred meter

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Answer:

efficiency = 5.4%

Explanation:

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now we will have

$W = Q_1 - Q_2$

so we will have

$\eta = 1 - \frac{Q_2}{Q_1}$

now we know that

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$\eta = 1 - \frac{273+6}{273+22}$

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3 years ago

What part of the roller coaster has the lowest potential energy?

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It’s 1 that would be the lowest point potential energy

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3 years ago

In a charming 19th-century hotel, an old-style elevator is connected to a counterweight by a cable that passes over a rotating d

melisa1 [442]

Answer:

2.38732 rpm

1.22625 rad/s²

163.292°

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration = $\frac{1}{8}g$

d = Diameter of wheel = 2 m

r = Radius of wheel = $\frac{d}{2}=\frac{2}{2}=1\ m$

v = Speed of elevator = 25 cm/s

Angular speed is given by

$\omega=\frac{v}{r}\\\Rightarrow \omega=\frac{0.25}{1}\\\Rightarrow \omega=0.25\ rad/s=0.25\times \frac{60}{2\pi}\\\Rightarrow \omega=2.38732\ rpm$

The angular speed of the wheel is 2.38732 rpm

Angular acceleration is given by

$\alpha=\frac{a}{r}\\\Rightarrow \alpha=\frac{\frac{1}{8}g}{r}\\\Rightarrow \alpha=\frac{\frac{1}{8}\times 9.81}{1}\\\Rightarrow \alpha=1.22625\ rad/s^2$

The angular acceleration of the wheel is 1.22625 rad/s²

Angular displacement is given by

$\theta=\frac{s}{r}\\\Rightarrow \theta=\frac{2.85}{1}\\\Rightarrow \theta=2.85\ rad=2.85\times \frac{360}{2\pi}\\ =163.292\ ^{\circ}$

The angle the disk turned when it has raised the elevator is 163.292°

4 0

3 years ago

A jogger accelerates from rest to 3.0 m/s in 2.0 s. A car accelerates from 38.0 to 41.0 m/s also in 2.0 s. (a) Find the accelera

iren [92.7K]

Answer:

(a) a₁: jogger acceleration= 1.5 m/s²

(b) a₂: car acceleration = 1.5 m/s²

(b) d= 76m : the car travels 76 meters longer than the jogger during the 2 seconds

Explanation:

we apply uniformly accelerated motion formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:

d:displacement in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Nomenclature

d₁: jogger displacement

t₁ : jogger time

v₀₁: jogger initial speed

vf₁: jogger final speed

a₁: jogger acceleration

d₂: car displacement

t₂ : car time

v₀₂: car initial speed

vf₂: car final speed

a₂: car acceleration

Data

v₀₁ = 0

vf₁ = 3 m/s

t₁ =2.0 s

v₀₂ = 38.0m/s

vf₂ = 41.0 m/s

t₂ = 2.0 s

Problem development

(a) Find the acceleration (magnitude only) of the jogger.

We apply the formula (1) for calculate acceleration :

vf₁= v₀₁+a₁*t₁

3 = 0 +(a₁)*(2)

a₁= (3)/(2)

a₁= 1.5 m/s²

(b) Determine the acceleration (magnitude only) of the car.

We apply the formula (1) for calculate acceleration :

vf₂= v₀₂+a₂*t₂

41 = 38 +(a₂)*(2)

a₂= (41 - 38)/(2)

a₂= 3 /2

a₂= 1.5 m/s²

We apply the formula (1) for calculate distance :

d₁= v₀₁*t₁+ (1/2)*a₁*t₁²= 0+ (1/2)*(1.5) *(2)² = 3 m

d₂= v₀₂*t₂+ (1/2)*a₂*t₂² =38*(2)+ (1/2)*(1.5) *(2)²= 79 m

d= 79 m-3 m

d= 76m : the car travels 76 meters longer than the jogger during the 2 seconds

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3 years ago

what two objects have stored energy? A. ball rolling on the ground B. a small rock sitting on top of a big rock C. a stretched r

Verizon [17]

The correct statements are:

B. a small rock sitting on top of a big rock

As the rock is at a height with respect to ground it has potential Energy

and

C. a stretched rubber band

A stretched rubber band has elastic potential energy

The others are actually moving and hence would consist of Kinetic energy. Potential energy is stored in objects that do not move and are stationary.

8 0

3 years ago

Read 2 more answers