Ask question

Ask question

All categories

3 years ago

13

Astronauts are testing the gravity on a new planet. A rock is dropped between two photogates that are 0.5 meters apart. The firs

t photogate reads a velocity of 1.2 m/s and the the second photogate reads a velocity of 4.3 m/s . What is the acceleration of gravity on this new planet?

1 answer:

KengaRu [80]3 years ago

5 0

Answer:

a = 17 m / s²

Explanation:

For this experiment that the astronauts are carrying out, the value of the relation of gravity is cosecant, therefore we can use the kinematic relations

v² = v₀² + 2a y

They indicate the initial speed 1.2 m / s the final speed 4.3 m / s and the distance remembered 0.5 m

we clear

a = (v² - v₀²) / 2y

we calculate

a = (4.3² -1.2²) / 2 0.5

a = 17 m / s²

this is the gravity of the new planet

You might be interested in

A sound wave takes 0.2 seconds to travel 306 meters.

chubhunter [2.5K]

Answer:

1530 m/s

Explanation:

Given that :

Speed = distance / time

Travel time take = 0.2 s

Distance covered = 306 metres

The speed of sound in the material :

Speed = distance / time = 306 m / 0.2

Speed = 1530 m/s

6 0

3 years ago

The motion of an electron is given by x(t)=pt3+qt2+r, with p = -2.3 m/s^3 ,q = +1.5 m/s^2 , and r = +9.0 m.A) Determine its velo

alexandr402 [8]

Answer:

$v(0)=0\\v(1)=-3.9\ m/s\\v(2)=-21.6\ m/s\\v(3)=-53.1\ m/s$

Explanation:

<u>Instant Velocity </u>

Given the position as a function of time x(t), the instant velocity is the derivative of the function:

$v(t)=x'(t)$

We are given the position as

$x(t)=-2.3t^3+1.5t^2+9$

The derivative of x is

$v(t)=x'(t)=-6.9t^2+3.0t$

A) Let's compute v(0)

$v(0)=-6.9(0)^2+3.0(0)=0$

B)

$v(1)=-6.9(1)^2+3.0(1)$

$v(1)=-3.9\ m/s$

C)

$v(2)=-6.9(2)^2+3.0(2)$

$v(2)=-21.6\ m/s$

D)

$v(3)=-6.9(3)^2+3.0(3)$

$v(3)=-53.1\ m/s$

4 0

3 years ago

In a paragraph of no less than five complete sentences, explain how the bubble gum experiment demonstrates the Law of Conservati

Marianna [84]

Answer:

The bubble gum experiment demonstrates the law of conservation of mass in that even though the bubble gum has lost mass, this loss in mass is not because some matter present in the gum has been destroyed, but it has changed form and has been removed from the gum.

Explanation:

In this science experiment, students investigate whether or not chewing gum should be considered eating. During the process of chewing the gum, the gum loses mass. The experiment is used to demonstrate the law of conservation of mass which states that matter can neither be created nor destroyed but may change from one form to another.

The loss in mass of the gum is due to the fact that the sugar present in the gum has changed form and has been removed from the gum. During the process of chewing the gum, the sugar in solid form present in the gum is dissolved in the saliva found in the mouth. The dissolved sugar is then swallowed and passes into the digestive tract for digestion. This shows that even though the bubble gum has lost mass, this loss in mass is not because some matter present in the gum has been destroyed, but it has changed form and has been removed from the gum, This demonstrates the law of conservation of mass.

4 0

3 years ago

A box (m = 20 kg) is sliding on a horizontal surface. it is connected to a massless hook by a light string passing over a massle

Varvara68 [4.7K]

mass of the box = 20 kg

force of friction on the box due to surface

$F_s = \mu_s N$

$F_s = 0.80 * 20 * 9.8$

$F_s = 156.8 N$

similarly kinetic friction on it

$F_k = \mu_k N$

$F_k = 0.30* 20 * 9.8$

$F_k = 58.8 N$

now the weight of the suspended block will be

$W = mg = 15*9.8 = 147 N$

so here the weight of the suspended block is less than the limiting friction on it

So here we will say that friction will counter balance the weight of the suspended block and it will not move at all

So acceleration of the box will be zero

5 0

4 years ago

At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is

Katyanochek1 [597]

Answer:

$4.47\ \text{km/h}$

Explanation:

$\dfrac{da}{dt}$ = Rate at which the distance between A and starting point of B is changing = -20 km/h

$\dfrac{db}{dt}$ = Rate at which the distance of B is changing = 15 km/h

$\dfrac{dc}{dt}$ = Rate at which the distance between A and B is changing

Time after which the rate at which the distance between A and B is changing is 4 hours

Distance covered by A in 4 hours = $20\times 4=80\ \text{km}$

a = Distance remaining to the start point of B = $110-80=30\ \text{km}$

b = Distance covered by B in 4 hours = $15\times 4=60\ \text{km}$

Distance between A and B after 4 hours

$c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}$

$c^2=a^2+b^2$

Differentiating with respect to time we get

$c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}$

The rate at which the distance between the ships is changing at 4 PM is $4.47\ \text{km/h}$ .

7 0

3 years ago