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den301095 [7]
3 years ago
10

If you want quick access to large amount of data, you would use a

Physics
1 answer:
RSB [31]3 years ago
6 0
You would use d. table

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Plane a travels at 900km/h and plane b travels at 250/5.which plane travels faster
vesna_86 [32]

Explanation:

We have,

Speed of plane a is 900 km/h

Plane b is moving at a rate of \dfrac{250\ km}{5\ h}=50\ km/h

It is required to find which plane is faster. To find which plane is faster, we need to compare their speeds.

Speed of a plane a is 900 km/h and that of plane b is 50 km/h. So, we can say that plane a is moving faster.

5 0
3 years ago
A car moves at 20 m s' for 15 minutes. Calculate the distance travelled
Firdavs [7]

Explanation:

Calculating this according to the m/s rate

Now solving the question

If the car goes 20 m/s mathematically we can generate it as

15 × 60 = 600 secs

If the car goes 20 meters per every second it means

The car will go 20 meters for 900 secs

Which is 900 ×20 = 18000m/s

=18000÷1000 = 1km

Therefore the answer is 18km

6 0
2 years ago
A balloon is rising vertically upwards at a velocity of 10m/s. When it is at a height of 45m from the ground, a parachute bails
harina [27]

(a) 30.9 m

Let's analyze the motion of the parachutist. Its vertical position above the ground is given by

y=h+ut+\frac{1}{2}gt^2

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t=3 s , we find the height of the parachutist when it opens the parachute:

y=45 m+(10 m/s)(3 s)+\frac{1}{2}(-9.8 m/s^2)(3 s)^2=30.9 m

(b) 44.1 m

Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by

y = h + ut

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

Substituting t = 3 s, we find

y = 45 m + (10 m/s)(3 s) = 75 m

So the distance between the balloon and the parachutist after 3 s is

d = 75 m - 30.9 m = 44.1 m

(c) 8.2 m/s downward

The velocity of the parachutist at the moment he opens the parachute is:

v = u +gt

where

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t = 3 s,

v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s

where the negative sign means it is downward

After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of

a =+5 m/s^2

where we put a positive sign since this time the acceleration is upward.

The total distance he still has to cover till the ground is

d = 30.9 m

So we can find the final velocity by using

v^2-u^2 = 2ad

where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:

a = -5 m/s^2

Solving for v,

v=\sqrt{u^2 +2ad}=\sqrt{(19.4 m/s)^2+2(-5 m/s^2)(30.9 m)}=8.2 m/s

(d) 5.24 s

We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using

a=\frac{v-u}{t}

where

a = -5 m/s^2 is the deceleration

v = 8.2 m/s is the final velocity

u = 19.4 m/s is the initial velocity

t is the time

Solving for t, we find

t=\frac{v-u}{a}=\frac{8.2 m/s-19.4 m/s}{-5 m/s^2}=2.24 s

And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is

t = 3 s + 2.24 s = 5.24 s

8 0
2 years ago
Why is the unit of velocity called derived unit ?​
navik [9.2K]

Answer:

Explanation:

The general consensus is that it's more “natural” to define distance (meter) and time (second) and as base units, and derive velocity a the ratio between them. ... The general consensus is that it's more “natural” to define distance (meter) and time (second) and as base units, and derive velocity a the ratio between them.

6 0
3 years ago
Friction is a force that always acts
insens350 [35]
I’m pretty sure it does most of the time ig
7 0
2 years ago
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