Does the planet exert a torque on the meteoroid with respect to the center of mass of the planet? Why or why not?

ListenA bicycle and its rider have a combined mass of 80. kilograms and a speed of 6.0 meters per second. What is the magnitude

Setler [38]

Answer:

a) $1.2\times 10^2\ N$

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

a = Acceleration

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{0-6}{4}\\\Rightarrow a=-1.5\ m/s^2$

The acceleration of the bicycle and rider is -1.5 m/s²

Force

$F=ma\\\Rightarrow F=80\times -1.5\\\Rightarrow F=-120\ N=-1.2\times 10^2\ N$

The magnitude of the average force needed to bring the bicycle and its rider to a stop is $1.2\times 10^2\ N$

3 0

3 years ago

A humanoid skeleton is found buried in the ashes of a volcano that erupted between 10,000 and 12,000 years ago. when scientists

Inga [223]

From reliable sources in the internet, the half-live of carbon-14 is given to be 5,730 years. In a span of 10,000 to 12,000 years, there are almost or little more than 2 half-lives. Thus, there should be
A(t) = A(0)(1/2)^t
where t is the number of half-lives, in this case 2. Thus, only about 1/4 of the original amount will be left.

5 0

3 years ago

Read 2 more answers

Does the horizontal distance d travelled by the ball depend on the height of release? If it does depend on the height, what is t

elena-s [515]

Answer:

Explanation:

Yes , the horizontal distance travelled by the ball will depend upon the height of release .

When a ball is thrown at some angle from a height , it has two components , the vertical component and horizontal component . The ball goes in horizontal direction due to its horizontal component . Its vertical component has no role to play . But the horizontal range covered by the body thrown

depends upon the duration of time in which it remains in air . The longer it remains in air , the greater distance it can cover horizontally .

Horizontal distance covered = t x horizontal velocity

If V be the velocity of throw and Vx be its horizontal component

Horizontal distance covered = t x Vx

Now t depends upon the height . If height rises , time of fall will increase so horizontal distance covered will increase .

If h be the height from which the body is thrown , Vy be the vertical upward component of initial velocity

from the relation

s = ut + 1/2 at²

h = - Vy t + 1/2 at²

As h increases , t will increase and therefore horizontal distance covered will increase. If the ball has only horizontal velocity initially , Vy = 0

h = 1/2 gt²

$t = \sqrt{\frac{2h}{g} }$

Horizontal distance covered = t x Vx

= $\sqrt{\frac{2h}{g} } \times V_x$

From this expression also

Horizontal distance covered is proportional to $\sqrt{h}$ .

7 0

3 years ago

Moving the probe 1 cm towards the non-grounded electrode changes the value the potential from about 0.90 V to about 1.2 V. Expla

svp [43]

Answer:

-30 N/C

Explanation:

Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m

Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V

Since E = -ΔV/Δx

substituting the values of the variables into the equation, we have

E = -ΔV/Δx

E = -0.30 V/0.01 m

E = -30 V/m

Since 1 V/m = 1 N/C.

E = -30 N/C

So, the average electric field is -30 N/C

6 0

3 years ago

N capacitors are connected in parallel to form a "capacitor circuit". The capacitance of first capacitor is C, second one is C/2

liberstina [14]

Answer:

2C

Explanation:

The equivalent capacitance of a parallel combination of capacitors is the sum of their capacitance.

So, if the capacitance of each capacitor is half the previous one, we have a geometric series with first term = C and rate = 0.5.

Using the formula for the sum of the infinite terms of a geometric series, we have:

Sum = First term / (1 - rate)

Sum = C / (1 - 0.5)

Sum = C / 0.5 = 2C

So the equivalent capacitance of this parallel connection is 2C.

5 0

3 years ago