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3 years ago

What do we use as a reference point to prove that we are always moving even when we are sitting down

2 answers:

4 0

Im not quite sure what you mean but do you mean us or the earth? If you mean us a reference is that our hearts are beating, we are breathing, we (may be) blinking, we are using our brains to do everything, etc.

mafiozo [28]3 years ago

3 0

This is a elementary question? in my opinion we could use stars. as the earth is in constant rotation, therefor we are technically still moving if we are sitting, just like being in a car, if that makes since

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The spaceship Intergalactic landed on the surface of the uninhabited Pink Planet, which orbits a rather average star in the dist

max2010maxim [7]

Answer:

Explanation:

Given

Length of the pendulum L = 1.44 m

Number of complete cycles of oscillation n = 1.10 x 10^2

total time of oscillation t = 2.00 x 10^2 s

The period of the T = n/t

T = (1.10 x 10^2)/(2.00 x 10^2) = 0.55 ^-s

The period of a pendulum is gotten as

T = $2\pi \sqrt{\frac{L}{g} }$

where g is the acceleration due to gravity

substituting values, we have

0.55 = $2\pi \sqrt{\frac{1.44}{g} }$

0.0875 = $\sqrt{\frac{1.44}{g} }$

squaring both sides of the equation, we have

7.656 x 10^-3 = 144/g

g = 144/(7.656 x 10^-3) = <em>18808.7 m/s^2</em>

3 0

3 years ago

A basketball of mass 0.23kg is thrown horizontally against a rigid vertical wall with a velocity of 20m/s. It rebounds with a ve

Anna11 [10]

Answer:

$8.1\:\mathrm{Ns}$

Explanation:

The impulse-momentum theorem gives the impulse on an object to be equal to the change in momentum of that object. Since mass is maintained, the change in momentum of the basketball is:

$\Delta p = m\Delta v$ , where $m$ is the mass of the basketball and $\Delta v$ is the change in velocity.

Since the basketball is changing direction, its total change in velocity is:

$\Delta v = 20-(-15)=35\:\mathrm{m/s}$ .

Therefore, the basketball's change in momentum is:

$\Delta p = m\Delta v = 0.23\cdot 35= 8.05=8.1\:\mathrm{kg\cdot m/s}$ .

Thus, the impulse on the basketball is $\fbox{$8.1\:\mathrm{Ns}$}$ (two significant figures).

7 0

3 years ago

A 13 500 N car traveling at 50.0 km/h rounds a curve of radius 2.00 × 102 m. Find the following: a. the centripetal acceleration

Afina-wow [57]

Answer:

a. 0.947 m/s^2

b. 1304.54 N

c. 0.0966

Explanation:

mass of car = 13500 N = 13500/9.8 = 1377.55 kg

velocity = 50 km/h = 50,000 m/h = 13.9 m/s

raidus = 204 m

a. centripetal acceleartion = v^2/r = 13.9^2/204 = 0.947 m/s^2

b. centripetal force = m*centripetal acceleration = 1377.55 * 0.947 = 1304.54 N

c. In order for the car to round the curve safely, static friction = centripetal force

static friction = coefficient of friction (mu) * mg = mu* 1377.55*9.8 = 13500mu

13500mu = 1304.54

mu = 1304.54/13500 = 0.0966

5 0

3 years ago

A 14,000-kg boxcar is coasting at 1.50 m/s along a horizontal track when it suddenly hits and couples with a stationary 10,000-k

ludmilkaskok [199]

Explanation:

It is given that,

Mass of a box car, $m_1=14000\ kg$

Initial speed of a box car, $u_1=1.5\ m/s$

Initial speed of another stationary car is 0 as it was stationary

Mass of another box car is 10,000 kg

Let V is the speed of the cars just after the collision. It is based on the concept of inelastic collision. So, using the conservation of linear momentum we get :

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\m_1u_1=(m_1+m_2)V\\\\V=\dfrac{m_1u_1}{(m_1+m_2)}\\\\V=\dfrac{14000\times 1.5}{(14000+10000)}\\\\V=0.875\ m/s$

So, the speed of the cars just after the collision is 0.875 m/s.

5 0

4 years ago

Can someone please help me with this question plz

Varvara68 [4.7K]

Section 2 and 5 so on this test C

8 0

3 years ago