answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
wolverine [178]
2 years ago
8

Two projectiles are launched with the same initial speed from the same location, one at a 30° angle and the other at a 60° angle

with the horizontal. They land at the same height at which they were launched. If air resistance is negligible, how do the projectiles’ respective maximum heights, H30 and H60 , and times in the air, T30 and T60 , compare with each other?
Physics
1 answer:
DanielleElmas [232]2 years ago
7 0

Answer:

Explanation:

Given

launch angle \theta _1=30^{\circ}

\theta _2=60^{\circ}

\theta _1 and \theta _2 are complimentary angles so range for both of them is same

H_{max}=\frac{u^2(\sin \theta )^2}{2g}

H_{30}=\frac{u^2(\sin 30)^2}{2g}

H_{30}=\frac{u^2}{8g}

time of flight =\frac{2u\sin \theta }{g}

t_{30}=\frac{u}{g}

For \theta =60^{\circ}

H_{60}=\frac{u^2(\sin 60)^2}{2g}

H_{60}=\frac{3u^2}{8g}

t_{60}=\frac{2u\sin 60}{g}=\frac{\sqrt{3}u}{g}

H_{60}=3\times H_{30}

t_{60}=\sqrt{3}\times t_{30}

You might be interested in
Write difference between upstroke and downstroke of water pump
lakkis [162]

Explanation:

Upstroke is a mechanism which helps to raise the plunger and downstroke helps to help lower the plunger. On the up-stroke of the plunger, the lower valve opens and the upper valve is closed. ... Whereas, on the downstroke, the lower valve closes and the upper one opens.

7 0
2 years ago
In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040.
lina2011 [118]

Answer:

42.11 years old

Explanation:

Given that:

In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040

To find her age we use:

\Delta t_m=\frac{\Delta t_s}{\sqrt{1-\frac{v^2}{c^2} } }\\

Δtm is  time interval for the observer stationary relative to the sequence of

events = 2040 - 2000 = 40 years

Δts is is the time interval for an observer moving with a speed v relative to the  sequence of event

v = velocity = 2.5 x 10^8 m/s

c = speed of light = 3 x 10^8 m/s

\Delta t_s=\Delta t_m}{\sqrt{1-\frac{v^2}{c^2} } }\\\Delta t_s=40\sqrt{1-\frac{(2.5*10^8)^2}{(3*10^8)^2}}\\\Delta t_s=22.11\ yr

Here age in 2000 is 20 year, therefore when she appear she would be 20 year + 22.11 year = 42.11 years old

7 0
2 years ago
Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50
defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

8 0
2 years ago
61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A
Mademuasel [1]

Answer:

Part a)

percentage = 21.3%

Part b)

percentage = 2.13 \times 10^{-5}%

Explanation:

As we know that total power used in the room is given as

P = P_1 + P_2 + P_3 + P_4

here we have

P_1 = (110)(3) = 330 W

P_2 = 100 W

P_3 = 60 W

P_4 = 3 W

P = 330 + 100 + 60 + 3

P = 493 W

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

110\times i = 493

i = 4.48 A

now resistance of transmission line

R = \frac{\rho L}{A}

R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}

R = 5.23 \ohm

now power loss in line is given as

P = i^2 R

P = (4.48)^2(5.23)

P = 105 W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{105}{493} \times 100

percentage = 21.3%

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

110 \times 10^3 (i ) = 493

i = 4.48\times 10^{-3} A

now power loss in line is given as

P = i^2 R

P = (4.48 \times 10^{-3})^2(5.23)

P = 1.05 \times 10^{-4} W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{1.05 \times 10^{-4}}{493} \times 100

percentage = 2.13 \times 10^{-5}%

6 0
2 years ago
A 1500 kg car is pushing a 4000 kg truck. The car and truck are accelerating at 2.0 m/s^2. Assuming that the frictional force on
Crazy boy [7]

To solve this problem we will use the Force equation according to the definition given in Newton's second law. There we have that the Force is equal to

F =ma

Where,

m = mass

a = acceleration

Our values are given as

m_1 = 1500 kg

m_2 = 4000 kg

a = 2.0 m/s^2

Considering that both mass are equal to one, we have that:

F = (m_1+m_2)* a

F = (1500 + 4000)(2.0)

F = 11000 N = 11kN

Therefore the truck exert a force on the car of 11kN

7 0
2 years ago
Other questions:
  • A ladder is leaning against a vertical wall, and both ends of the ladder are at the point of slipping. The coefficient of fricti
    10·1 answer
  • Which type of listening response includes the use of head nods, facial expressions, and short utterances such as "uh-huh" that s
    8·1 answer
  • You throw a beanbag in the air and catch it 2.2 s later at the same place at which you threw it. How high did it go? What was th
    9·1 answer
  • An electric pump rated 1.5 KW lifts 200kg of water through a vertical height of 6m in 10 secs: way is the efficiency of the pump
    13·1 answer
  • Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.4 kg gibbon
    15·1 answer
  • A hiker walks due east for a distance of 25.5 km from her base camp. On the second day, she walks 41.0 km northwest till she dis
    7·1 answer
  • The atmosphere pressure can support mercury in a tube, which the upper end is closed, up to 0.76 meter. If the mercury is replac
    11·1 answer
  • The length of a wooden rod is 25.5 cm. What is this length in:<br>(a) millimetres?<br>(b) metres?​
    5·2 answers
  • A projectile is launched at an angle of 60° from the horizontal and at a velocity of
    15·1 answer
  • Venn diagrams are used for comparing and contrasting topics. The overlapping sections show characteristics that the topics have
    13·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!