Ask question

Ask question

All categories

3 years ago

11

A pitched ball is hit by a batter at a 45degrees angle and just clears the outfield fence, 98m away. Assume that the fence is at

the same height as the pitch and find the velocity of the ball when it left the bat. Neglect air resistance.

1 answer:

Alex17521 [72]3 years ago

8 0

The range of a projectile can be found directly using:
R = (v²sin2∅) / g
v = √((98 x 9.81)/(sin(90)))
v = 31.0 m/s

You might be interested in

Why couldnt mendeleev organize the entire table during his research

NARA [144]

Cause he left out the noble gases out of the periodic table for one good reason, 1: He did not know them

6 0

3 years ago

Read 2 more answers

A 4.0-m-diameter playground merry-go-round, with a moment of inertia of

HACTEHA [7]

Answer:

$7.1$ ms⁻¹

Explanation:

$d$ = diameter of merry-go-round = 4 m

$r$ = radius of merry-go-round = $\frac{d}{2}$ = $\frac{4}{2}$ = 2 m

$I$ = moment of inertia = 500 kgm²

$w_{i}$ = angular velocity of merry-go-round before ryan jumps = 2.0 rad/s

$w_{f}$ = angular velocity of merry-go-round after ryan jumps = 0 rad/s

$v$ = velocity of ryan before jumping onto the merry-go-round

$m$ = mass of ryan = 70 kg

Using conservation of angular momentum

$Iw_{i} - m v r = (I + mr^{2})w_{f}$

$(500)(2.0) - (70) v (2) = (I + mr^{2})(0)$

$1000 = 140 v$

$v = 7.1$ ms⁻¹

5 0

3 years ago

The force involved in falling of an apple from a tree is known as? (a) Magnetic force (b) electrostatic force (c) Contact force

musickatia [10]

Answer:

gravitational force

...........

6 0

3 years ago

Only 5 questions plz answer.

Masja [62]

Question 18: a
question 19: b
question 20: c

6 0

3 years ago

Read 2 more answers

Consider a double slit experiment in the air. The wavelength of light is 500nm light, the screen is 1m away and two adjacent bri

Bess [88]

Answer: $0.75\ cm$

Explanation:

Given

Wavelength of light $\lambda=500\ nm$

Screen is $D=1\ m$ away

Distance between two adjacent bright fringe is $\Delta y=\dfrac{\lambda D}{d}$

When same experiment done in water, wavelength reduce to $\dfrac{\lambda }{\mu}$

So, the distance between the two adjacent bright fringe is $\Delta y'=\dfrac{\lambda D}{\mu d}$

Keeping other factor same, distance becomes

$\Rightarrow \dfrac{1}{\frac{4}{3}}=\dfrac{3}{4}\quad \text{Refractive index of water is }\dfrac{4}{3}\\\\\Rightarrow 0.75\ cm$

3 0

3 years ago