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3 years ago

15

A ball is thrown into the air at an angle. What is the instantaneous velocity of this ball in the vertical direction at the top

of its flight?

1 answer:

Margarita [4]3 years ago

3 0

Answer:

0 m/s

Explanation:

At the instant the ball reaches the top of its flight, it isn't moving up or down. So the vertical component of the instantaneous velocity is 0 m/s.

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Answer: a) speed = 3.45 × 10^-2 m/s

b) speed = 1.38 × 10^-1 m/s

Explanation: shown in the attachment

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With what speed must a ball be thrown directly upward so that it remains in the air for 10 seconds?

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Answer:

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Explanation:

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3 years ago

a train is moving with an initial velocity of 30 m/s, the brakes are applied so as to produce a uniform acceleration of -1.5 m/s

Pepsi [2]

Answer:

$\boxed{\sf Time \ in \ which \ train \ will \ come \ to \ rest = 20 \ sec}$

Given:

Initial velocity (u) = 30 m/s

Final speed (v) = 0 m/s

Acceleration (a) = - 1.5 m/,s²

To Find:

Time in which train will come to rest (t).

Explanation:

$\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ a: \\ \sf \implies 0 = 30 + ( - 1.5)(t) \\ \sf \implies 0 = 30 - 1.5(t) \\ \sf \implies 30 - 1.5(t) = 0 \\ \\ \sf Subtract \: 30 \: from \: both \: sides: \\ \sf \implies (30 - \boxed{ \sf 30}) - 1.5(t) = \boxed{ \sf - 30} \\ \\ \sf 30 - 30 = 0 : \\ \sf \implies - 1.5(t) = - 30 \\ \\ \sf Divide \: both \: sides \: of \: - 1.5(t) = - 30 \: by \: - 1.5 : \\ \sf \implies \frac{ - 1.5(t)}{ \boxed{ \sf - 1.5}} = \frac{ - 30}{ \boxed{ \sf -1.5 }} \\ \\ \sf \frac{ \cancel{ \sf 1.5}}{\cancel{ \sf 1.5}} = 1 : \\ \sf \implies t = \frac{ - 30}{ - 1.5} \\ \\ \sf \frac{ - 30}{ - 1.5} = \frac{\cancel{ \sf 1.5} \times 20}{\cancel{ \sf 1.5}} = 20 : \\ \sf \implies t = 20 \: sec$

So,

Time in which train will come to rest = 20 seconds

4 0

3 years ago

What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction

Sergio039 [100]

Answer:

The frequency is 302.05 Hz.

Explanation:

Given that,

Speed = 18.0 m/s

Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .

We need to calculate the frequency

Using formula of frequency

$f'=f(\dfrac{v+v_{p}}{v-v_{s}})$

Where, f = frequency

v = speed of sound

$v_{p}$ = speed of passenger

$v_{s}$ = speed of source

Put the value into the formula

$f'=262\times(\dfrac{344+18}{344-30})$

$f'=302.05\ Hz$

Hence, The frequency is 302.05 Hz.

7 0

3 years ago