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BlackZzzverrR [31]

3 years ago

9

A pitched ball is hit by a batter at a 47◦ angle. It just clears the outfield fence, 98 m away. The acceleration of gravity is 9

.8 m/s 2 . Find the velocity of the ball when it left the bat. Assume the fence is the same height as the pitch. Answer in units of m/s.

1 answer:

musickatia [10]3 years ago

5 0

Answer:u=31.02 m/s

Explanation:

Given

launch angle of ball $\theta =47^{\circ}$

Range of ball $R=98 m$

considering ball to be a Projectile

Range of Projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

Where $u=initial\ Velocity$

$\theta =launch\ angle$

$g=acceleration\ due\ to\ gravity$

$98=\frac{u^2\sin 2(47)}{9.8}$

$98\times 9.8=u^2\sin (94)$

$u^2=\frac{960.4}{sin (94)}$

$u^2=962.74$

$u=31.02 m/s$

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Answer:

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5 0

3 years ago

Read 2 more answers

How to prove formula for volume of a sphere ?

atroni [7]

Answer:

The volume of radius is $\frac{4}{3}$ × π × radius³ Proved

Explanation:

Given as :

We know that volume of sphere is v = $\frac{4}{3}$ × π × radius³

Or, v = $\frac{4}{3}$ × π × r³

Let prove the volume of sphere

So, From the figure of sphere

At the height of z , there is shaded disk with radius x

Let Find the area of triangle with side x , z , r

<u>From Pythagorean theorem</u>

x² + z² = r²

Or, x² = r² - z²

Or, x = $\sqrt{r^{2}-z^{2} }$

Now, Area of shaded disk = Area = π × x²

Where x is the radius of disk

Or, Area of shaded disk = π × ( $\sqrt{r^{2}-z^{2} }$ ) ²

∴ Area of shaded disk = π × (r² - z²)

Again

<u>If we calculate the area of all horizontal disk, we can get the volume of sphere</u>

So, we simply integrate the area of all disk from - r to + r

i.e volume = $\int_{-r}^{r} \Pi(r^{2}-z^{2} )dz$

Or, v = $\int_{-r}^{r} \Pi r^{2}dz$ - $\int_{-r}^{r} \Pi z^{2}dz$

Or, v = π r² (r + r) - π $\frac{r^{3} -(-r)^{3})}{3}$

Or, v = π r² (r + r) - π $\frac{2r^{3}}{3}$

Or, v = 2πr³ - π $\frac{2r^{3}}{3}$

Or, v = 2πr³ ( $\frac{3-1}{3}$ )

Or, v = 2πr³ × $\frac{2}{3}$

∴ v = $\frac{4}{3}$ × π × r³

Hence, The volume of radius is $\frac{4}{3}$ × π × radius³ Proved . Answer

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3 years ago

Please help me do that.

Anika [276]

Answer:

a) 22.2 N

Explanation:

To find load effort. Identify load, efficiency and velocity ratio (VR)

Here given:

Load = 100 N

Efficiency = 75%

VR (number of pulleys) = 6

Effort :

$\rightarrow \sf \dfrac{MA}{VR} = \dfrac{Load \ \div \ efficiency }{VR} = \dfrac{100 \div 75 \% }{6} = 22.22 \ N$

3 0

3 years ago

A cell of e.m.f 1.5 v and internal resistance 2.5 ohm is connected in series with an ammeter of resistance 0.5 ohm and a resisto

sergejj [24]

Answer:

The current in the circuit is 0.15 Ampere

Explanation:

The given parameters of the cell are;

The electromotive force (e.m.f.) of the cell, E = 1.5 V

The resistance of the cell, r = 2.5 ohm

The resistance of the ammeter = 0.5 ohm

The resistance of the resistor = 7.0 ohm

The formula for the e.m.f., E of a cell is given as follows;

e.m.f. E = I·(R + r)

Where;

I = The current in the circuit

R = The sum of the resistances in the circuit = 7.0 Ω + 0.5 Ω + 2.5 Ω = 10 Ω

Therefore, we have;

$The \ current \ in \ the \ circuit, \ I = \dfrac{E}{R + r}$

Substituting the known values, gives;

$I = \dfrac{1.5 \ V}{7 \ \Omega + 0.5 \ \Omega + 2.5 \ \Omega} = \dfrac{1.5 \ V}{10 \ \Omega} = 0.15 \ A$

The current in the circuit, I = 0.15 Ampere.

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3 years ago

A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are

Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of $x= 5t^3+1$

Coordinate of position of $y=3t^2+2$

Time = 2.00 s

We need to calculate the acceleration

$a = \dfrac{d^2x}{dt^2}$

For x coordinates

$x=5t^3+1$

On differentiate w.r.to t

$\dfrac{dx}{dt}=15t^2+0$

On differentiate again w.r.to t

$\dfrac{d^2x}{dt^2}=30t$

The acceleration in x axis at 2 sec

$a = 60i$

For y coordinates

$y=3t^2+2$

On differentiate w.r.to t

$\dfrac{dy}{dt}=6t+0$

On differentiate again w.r.to t

$\dfrac{d^2y}{dt^2}=6$

The acceleration in y axis at 2 sec

$a = 6j$

The acceleration is

$a=60i+6j$

We need to calculate the net force

$F = ma$

$F = 3.00\times(60i+6j)$

$F=180i+18j$

The magnitude of the force

$|F|=\sqrt{(180)^2+(18)^2}$

$|F|=180.89\ N$

Hence, The net force acting on this object is 180.89 N.

3 0

4 years ago