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irakobra [83]
2 years ago
15

A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

l direction with a variable speed and from a variable vertical position and a fixed horizontal position x = 0. The robot is calibrated by adjusting the speed at which the sphere is launched and the height of the robot's sphere launcher. Depending on where the spheres land on the ground, students earn points based on the accuracy of the robot. The robot is calibrated so that when the spheres are launched from a vertical position y = H and speed vo, they consistently land on the ground on a target that is at a position x = D. Positive directions for vector quantities are indicated in the figure. a) On the axes below, sketch the graphs of the horizontal and vertical components of the sphere's velocity v as a function of time t between t = 0, when the sphere is launched and t =T. when the sphere hits the target. Label t =T for the horizontal component of the sphere's velocity and the vertical component of the sphere's velocity,​
Physics
1 answer:
Dvinal [7]2 years ago
8 0
Nobody will do that for 5 points loll
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It has been proposed that extending a long conducting wire from a spacecraft (a "tether") could be used for a variety of applica
denis23 [38]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The angle between shuttle's velocity and the Earth's field is  \theta =   24.2^o

Explanation:

From the question we are told that

     The length of eire let out is  L = 250 \ m

      The emf generated is \epsilon = 40 V

      The earth magnetic field is B = 5.0 *10^{-5} T

     The speed of the shuttle and tether is v =  7.80 * 10^3 \  m/s

The emf generated is mathematically represented as

                             \epsilon = L\ v\ B\ sin \ \theta

making \theta  the subject of the formula

                        \theta =   sin ^{-1}[ \frac{\epsilon}{L  * B  *v} ]

substituting values

                        \theta =   sin ^{-1}[ \frac{40}{250  * (5*10^{-5})  *(7.80 *10^{3})} ]

                        \theta =   24.2^o

6 0
2 years ago
On a warm summer day (31 ∘c), it takes 4.60 s for an echo to return from a cliff across a lake. on a winter day, it takes 5.00 s
xenn [34]
The question is missing, but I guess the problem is asking for the distance between the cliff and the source of the sound.

First of all, we need to calculate the speed of sound at temperature of T=31^{\circ}C:
v=(331+0.60 T) m/s = (331+0.6 \cdot 31) m/s = 349.6 m/s

The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):
v= \frac{2L}{t}
where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
L= \frac{vt}{2}= \frac{(349.6 m/s)/4.60 s)}{2}=  804.1 m
6 0
2 years ago
Karyotypes are done by matching up _____________________________ so that they are paired up. Question 11 options:
Rufina [12.5K]

Homologous Chromosomes

6 0
2 years ago
Read 2 more answers
An object has a mass of 8.00kg. What is the gravitational force on the object by the earth
Firdavs [7]
78.4 F because you do 8.00 muliplyed by 9.8
6 0
2 years ago
Bill leaves his 60 W desk lamp on every day, including weekends, for eight hours. After one month (30 days), how much total ener
maxonik [38]

' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.

That's all the physics we need to know to answer this question.
The rest is just arithmetic.

(60 joules/sec) · (30 days) · (8 hours/day) · (3600 sec/hour)

= (60 · 30 · 8 · 3600) (joule · day · hour · sec) / (sec · day · hour)

= 51,840,000 joules
__________________________________

Wait a minute !  Hold up !  Hee haw !  Whoa ! 
Excuse me.  That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's

(60 watts) · (30 days) · (8 hours/day) · (1 kW/1,000 watts)

= (60 · 30 · 8 · 1 / 1,000) (watt · day · hour · kW / day · watt)

= 14.4 kW·hour

Rounded to the nearest whole number:

14 kWh

7 0
2 years ago
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