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Artist 52 [7]
3 years ago
13

PLEASE HELP. Really important

Physics
1 answer:
Kruka [31]3 years ago
3 0

Explanation:

2) C would need the least effort, because the longer the effort distance, the least the effort applied.

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If Scoobie could drive a Jetson's flying car at a constant speed of 450.0 km/hr across oceans and space, approximately how long
vladimir2022 [97]

Answer:

The value is t = 3.6 \  days

Explanation:

From the question we are told that

The speed is v  =  450.0 km/h

The radius of the earth is R =  6200 \  km

Generally the circumfernce of the earth is mathematically evaluated as

C =  2\pi  R

=> C =   2 * 3.142 *   6200

=> C =  38960.8 \ km

Generally the time taken is mathematically represented as

t =  \frac{38960.8}{450.0}

         t = 86.6 \  hr

Converting to days

         t = \frac{86.6}{24}

=>       t = 3.6 \  days

7 0
3 years ago
jim makes an electromagnet by wrapping a copper wire around a nail and then connecting it to a battery. if he brings the end of
SIZIF [17.4K]
The nail will have the ability to pull the pile of paperclips in a magnetic field. Meaning the nail will work as if it is a magnet

7 0
3 years ago
Read 2 more answers
The position-time equation for a certain train is xf=2.8m+(8.1m/s)t+(2.9m/s^2)t^2.
mash [69]
The formula without numbers is
xf = xi + vxi(t) +  \frac{1}{2} (a) {t}^{2}
do the initial velocity would be
8.1 m/s
and the acceleration would be
2.9 m/s^2
3 0
3 years ago
a bullet is fired horizontally with a speed of 524 m/s from a height of 22m above the ground. calculate where it will hit the gr
lara31 [8.8K]

Given data:

* The height of the bullet is 22 m.

* The speed of bullet in the horizontal direction is 524 m/s.

Solution:

By the kinematics equation, the time taken by the bullet to reach the ground is,

h=u_yt+\frac{1}{2}gt^2

where u_y is the vertical velocity component, t is the time taken to reach the ground, g is the acceleration due to gravity, and h is the height of the bullet,

Substituting the known values,

\begin{gathered} 22=0+\frac{1}{2}\times9.8\times t^2 \\ t^2=\frac{22\times2}{9.8} \\ t^2=4.49 \\ t=2.12\text{ s} \end{gathered}

Thus, the time taken by the bullet to reach the ground is 2.12 seconds.

By the kinematics equation for the horizontal motion, the horizontal range of the bullet is,

R=u_xt+\frac{1}{2}at^2

where u_x is the horizontal component of the velocity, a is the acceleration along the horizontal direction, t is the time taken to reach the ground and R is the horizontal range,

Substituting the known values,

\begin{gathered} R=524\times2.12+0 \\ R=1110.9\text{ m} \end{gathered}

Thus, the horizontal range of the bullet is 1110.9 meters.

Hence, the bullet hit the ground at 1110.9 meters.

8 0
1 year ago
If you travel from Yakima to Ellensburg (Yakima to Ellensburg is 50 miles) with a speed of 60 miles/hour for half of the
koban [17]

Answer:

\displaystyle \frac{480}{7}\approx 68.6\; \rm mph.

Explanation:

The average speed of an object is equal to total distance over total time.

  • Distance traveled: \rm 50 \; mi.

How much time is taken? This trip is divided into two halves, each of distance \displaystyle \frac{50}{2} = 25\;\rm mi.

Time spent on the first half of the trip:

\displaystyle t_1 = \frac{s_1}{v_1} = \frac{25}{60} = \frac{5}{12}\; \rm hours.

Similarly, time spent on the second half of the trip:

\displaystyle t_2 = \frac{s_2}{v_2} = \frac{25}{80} = \frac{5}{16}\; \rm hours.

In total:

\displaystyle \frac{5}{12} + \frac{5}{16} = \frac{35}{48} \; \rm hours.

Average speed:

\begin{aligned} \text{Average speed} &= \frac{\text{Total Distance}}{\text{Total Time}}\\ &= 50 \left/\frac{35}{48}\right.\\ &= 50 \cdot \frac{48}{35} \\&= \frac{480}{7}\approx 68.6\; \rm mph \end{aligned}.

This value turned out to be slightly different from the average of the speed during the two halves of the journey. The reason is that the object traveled at each speed for a different amount of time. It spent more time at the slower speed, which gives that speed a greater weight in the average. That explains why the average speed is closer to \rm 60\; mph rather than \rm 80\; mph.

3 0
3 years ago
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