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2 years ago

Describe the motion of a person who is standing on the earths surface

Physics

1 answer:

oee [108]2 years ago

8 0

They are rotating because the earth never stops rotating.

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Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

kumpel [21]

Answer: $\frac{D_A}{v_B-v_A}$

Explanation:

Given

car A had a head start of $D_A$

and it starts at x=0 and t=0

Car B has to travel a distance of $D_A and d_a$

where $d_a$ is the distance travel by car A in time t

distance travel by car A is

$d_a=v_A\times t$

For car B with speed $v_B$

$d_B=D_A+d_a$

$v_B\times t=D_A+v_A\times t$

$t=\frac{D_A}{v_B-v_A}$

7 0

3 years ago

At what angle of projectile range of maximum

Agata [3.3K]

Answer:

$45°$

Explanation:

The textbooks say that the maximum range for projectile motion (with no air resistance) is 45 degrees.

7 0

3 years ago

An Airbus A380 airliner can takeoff when its speed reaches 80 m/s. Suppose its engines together can produce an acceleration of 3

Tomtit [17]

Answer:

(a). The time is 26.67 sec.

(b). The distance traveled during this period is 1066.9 m.

Explanation:

Given that,

Speed = 80 m/s

Acceleration = 3 m/s

Initial velocity = 0

(a). We need to calculate the time

Using equation of motion

$v = u+at$

$t = \dfrac{v-u}{a}$

Put the value into the formula

$t = \dfrac{80-0}{3}$

$t =26.67\ sec$

The time is 26.67 sec.

(b). We need to calculate the distance traveled during this period

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s = 0+\dfrac{1}{2}\times3\times(26.67)^2$

$s =1066.9\ m$

The distance traveled during this period is 1066.9 m.

Hence, This is the required solution.

7 0

3 years ago

If this decay has half-life of 2 years, how many years would it take for 10.8 g Protactinium-231 to remain given an initial mass

sergij07 [2.7K]

Answer:

Time = 6 years

Explanation:

First, we will calculate the no. of half life periods required to reduce the mass of Protactinium to the given value:

$m' = \frac{m}{2^{n} } \\\\2^n = \frac{m}{m'}$

where,

n = no. of half-life periods = ?

m = initial mass = 86.3 g

m' = remaining mass = 10.8 g

Therefore,

$2^n = \frac{86.3\ g}{10.8\ g}\\\\2^n = 8\\2^n = 2^3$

Since the bases are the same. Therefore equating powers:

n = 3

Now we calculate the time:

$Time = (n)(Half-Life)\\Time =(3)(2\ years)$

<u>Time = 6 years</u>

3 0

3 years ago

______ is the limiting factor that can keep some people from exercising if they can’t afford to purchase certain types of exerci

Karolina [17]

Answer:

I think it is Location.

Explanation:

If im wrong im sorry

5 0

3 years ago

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