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4 years ago

Challenge Question (This is supposed to be hard, think critically and take it one step at a time):

Physics

1 answer:

aivan3 [116]4 years ago

8 0

Answer:

You must travel at an average speed of 67.06 m/s to be on time to the lecture.

Explanation:

From the question, the total distance from your house to the science lecture is 75 miles.

Also, you get halfway there before you stop for a gas, that is, you have covered half of 75 miles, which 37.5 miles and you also have to cover 37.5 miles to get to the science lecture.

After filling up, you only have 15 minutes before the lecture starts,

To determine how fast you must drive to be on time to the lecture,

we will determine the average speed you need to travel.

From

Average speed = Distance / Time

Distance = 37.5 miles (Convert to meters)

(NOTE: 1 mile = 1609.344 meters)

Hence, 37.5 miles = 37.5 × 1609.344 miles = 60350.4 meters

∴ Distance = 60350.4 meters

Time = 15 minutes (Convert to seconds)

(NOTE: 1 minute = 60 seconds)

Hence, 15 minutes = 15 × 60 seconds = 900 seconds

Now, from

Average speed = Distance / Time

Average speed = 60350.4 m / 900 s

Average speed = 67.06 m/s

Hence, you must travel at an average speed of 67.06 m/s to be on time to the lecture.

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ListenA person on a ledge throws a ball vertically downward, striking the ground below the ledge with 200 joules of kinetic ener

polet [3.4K]

Answer:

A. 200 J

Explanation:

The initial kinetic energy depends on the initial speed, while the gravitational potential energy depends on the height, both balls are thrown with the same initial speed and from the same height. Therefore, due to the law of conservation of energy, the balls must have the same mechanical energy (the sum of both energies) when both impact the ground. Since the potential energy is zero at this point, its final kinetic energy must also be the same.

8 0

3 years ago

What is the highest pHoney bees beat their wings, making a buzzing sound at a frequency of 2.3 × 102 hertz. What is the period o

VMariaS [17]

The period of the wave is 4.35 ms. The sound waves are called longitudinal waves

Explanation:

The period of a wave is related to its frequency by the equation:

$T=\frac{1}{f}$

where

T is the period

f is the frequency

For the bee in this problem, the frequency of the sound wave emitted by it is

$f=2.3 \cdot 10^2 Hz = 230 Hz$

Therefore, the period of the sound wave is

$T=\frac{1}{230}=4.35\cdot 10^{-3}s = 4.35 ms$

The sound wave is a type of wave called longitudinal wave. In longitudinal waves, the oscillation of the medium occurs in a direction parallel to the direction of motion of the wave: therefore in a sound wave, the particle of the medium (air, in this case) oscillate back and forth along the direction of propagation of the wave, forming alternating areas of higher density of particles (called compressions) and of lower density of particle (called rarefactions).

The other type of wave, instead, is called transverse wave. In a transverse wave, the oscillation of the wave occurs in a direction perpendicular to the direction of motion of the wave. An example of transverse waves are the electromagnetic waves, which consists of electric field and magnetic fields that vibrate in a plane perpendicular to the direction of motion of the wave itself.

Learn more about waves:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

5 0

3 years ago

A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the

Ierofanga [76]

Answer:

The discharge rate is $Q = 0.0192 \ m^3 /s$

Explanation:

From the question we are told that

The diameter is $d = 60 \ mm = 0.06 \ m$

The head is $h = 6 \ m$

The coefficient of contraction is $Cc = 0.68$

The coefficient of velocity is $Cv = 0.92$

The radius is mathematically evaluated as

$r = \frac{d}{2}$

substituting values

$r = \frac{ 0.06 }{2}$

$r = 0.03 \ m$

The area is mathematically represented as

$A = \pi r^2$

substituting values

$A = 3.142 * (0.03)^2$

$A = 0.00283 \ m^2$

The discharge rate is mathematically represented as

$Q = Cv *Cc * A * \sqrt{ 2 * g * h}$

substituting values

$Q = 0.68 * 0.92* 0.00283 * \sqrt{ 2 * 9.8 * 6}$

$Q = 0.0192 \ m^3 /s$

6 0

3 years ago

A 80-kg person sits on a 2.7kg chair. Each leg of the chair makes contact with the floor on a circle that is 1.3 cm in diameter.

Rashid [163]

Answer:

need help with this 2

Explanation:

pls hellp

6 0

3 years ago

Read 2 more answers

Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot

Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² = 2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ = m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁ = 4.427 - 2v₂ ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂ = 8.854

v₂ = 8.854 / 3

v₂ = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁ = - 1.477 m/s

v₁ = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

$mgh_{max} = \frac{1}{2}mv_{max}^2$

(a) The maximum height achieved by the first ball (v₁ = 1.477 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max} = 0.11 \ m$

(b) The maximum height achieved by the second ball (v₂ = 2.95 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max} = 0.44 \ m$

6 0

3 years ago