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3 years ago

What times what equals 400

1 answer:

7 0

If you're willing to consider fractions or decimals,
then there are an infinite number of answers.
Like (2.5 x 160), and (15 x 26-2/3).

If you want to stick to only whole numbers,
then these 8 combinations do:

1, 400
2, 200
4, 100
5, 80
8, 50
10, 40
16, 25
20, 20

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A bumblebee darts past at 3 m/s. The frequency of the hum made by its wings is 152 Hz. Assume the speed of sound to be 342 m/s.

Veronika [31]

It'll be 152 Hz at the exact instant the bumblebee
is right at the tip of your nose, on his way past you.

Before he gets there, while he's coming at you,
he sounds like a frequency higher than 152 Hz.

After he passes by, and is going away from you,
he sounds like a frequency lower than 152 Hz.

8 0

3 years ago

Read 2 more answers

2.) The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his/her he

Ratling [72]

Answer:

The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s

Explanation:

The given parameters of the ball are;

The initial speed of the ball = 15 m/s

The direction in which the ball is launched = 50° above the horizontal

The location of the other tennis player when the ball is launched = 10 m from the ball

The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched

The height at which the ball is hit back = 2.1 m above the height from which the ball is launched

The vertical position, 'y', at time, 't', of a projectile motion is given as follows;

y = (u·sinθ)·t - 1/2·g·t²

When y = 2.1 m, we have;

2.1 = (15·sin(50°))·t - 1/2·9.8·t²

∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0

Solving with the aid of a graphing calculator function, we get;

t = 0.199776187257 s or t = 2.14525782198 s

Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds

The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;

x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m

The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m

Therefore, we have;

The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s

The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m

The minimum average speed the other player has to move with, $v_s$ = d/t₂

∴ $v_s$ = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s

The minimum average speed the opponent must move so that he is in position to hit the ball, $v_s$ ≈ 5.79 m/s.

5 0

3 years ago

(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W

Kitty [74]

Answer:

The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

$D\sin\theta=n\lambda$

$D=\dfrac{n\lambda}{\sin\theta}$

Where, D = width of slit

$\lambda$ = wavelength

Put the value into the formula

$D=\dfrac{n\lambda}{\sin\theta}$

Here, $\sin\theta$ should be maximum.

So. maximum value of $\sin\theta$ is 1

Put the value into the formula

$D=\dfrac{1\times\lambda}{1}$

$D=\lambda$

(b). If the minimum number is 50

Then, the width is

$D=\dfrac{50\times\lambda}{1}$

$D=50\lambda$

(c). If the minimum number is 1000

Then, the width is

$D=\dfrac{1000\times\lambda}{1}$

$D=1000\lambda$

Hence, The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

4 0

3 years ago

1.A body of certain mass is kept at a height of 15m from the ground, taking the value of g as 10m/s find out the mass of the bod

ehidna [41]

Answer:

1.#potential energy = PE, m = mass in kg, g = force of gravity, h= vertical height above the ground. ** means to the power of ie exponent. * means multiply.

PE = mgh

300 = m(10)(15)

m = 300/(10)(15)

m= 2kg

2. KE = 1/2 mv**2

= 1/2(50)(50)**2

= 2500 joules

Explanation

Is as in solution

4 0

3 years ago

A satellite is in a circular orbit 21000 km above the Earth’s surface; i.e., it moves on a circular path under the influence of

mina [271]

Answer:

(orbital speed of the satellite) V₀ = 3.818 km

Time (t) = 4.5 × 10⁴s

Explanation:

Given that:

The radius of the Earth is 6.37 × 10⁶ m; &

the acceleration of gravity at the satellite’s altitude is 0.532655 m/s

We can calculate the orbital speed of the satellite by using the formula:

Orbital Speed (V₀) = √(r × g)

radius of the orbit (r) = 21000 km + 6.37 × 10⁶ m

= (2.1 × 10⁷ + 6.37 × 10⁶) m

= 27370000

= 2.737 × 10⁷m

Orbital Speed (V₀) = √(r × g)

Orbital Speed (V₀) = √(2.737 × 10⁷ × 0.532655 )

= 3818.215

= 3.818 × 10³

= 3.818 Km

To find the time it takes to complete one orbit around the Earth; we use the formula:

Time (t) = 2 π × $\frac{r}{V_o}$

= 2 × 3.14 × $\frac{2.737*10^7}{3.818*10^3}$

= 45019.28

= 4.5 × 10 ⁴ s

6 0

3 years ago