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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.
Physics
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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For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser poin
Art [367]

Answer:

width of slit(a)≅ 0.1mm

Explanation:

Wave length of laser pointer =λ = 685 nm

Distance between screen and slit = L = 5.5 m

Width of bright band = W=8.0cm=0.08m

width of slit=a

recall the formula;

W=(2λL)/a

a=2λL/W

a=(2 *685*10⁻⁹*5.5m)/0.08m

a=7535*10⁻⁹/0.08

a=94187.5 *10⁻⁹

a=0.0000941875m

a=0.0941875mm

a≅0.1mm

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

5 0
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.

6 0
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
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