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3 years ago

Distance = 6 km south 60 minutes What was the average velocity

Physics

2 answers:

shutvik [7]3 years ago

8 0

For this case we have to by definition:

$v = \frac {x} {t}$

Where:

v: Velocity

x: Distance traveled

t: Time

According to the data we have:

$x = 6 \ km = 6000 \ m\\t = 60 \ min = 3600 \ s$

So:

$v = \frac {6} {60} = 0.1 \frac {km} {min}$

Equivalent to:

$v = \frac {6000} {3600} = 1.67 \frac {m} {s}$

Answer:

$0.1 \frac {km} {min}$

blsea [12.9K]3 years ago

3 0

v = x/t

v = average velocity, x = displacement, t = elapsed time

Given values:

x = 6km south, t = 60min

Plug in and solve for v:

v = 6/60

v = 0.1km/min south

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Why don’t we feel the gravitational force of a large object such as a skyscraper semi-truck?

Kobotan [32]

Answer:

Se the explanation below

Explanation:

We do not feel these forces of these bodies, because they are very small compared to the force of Earth's attraction. Although its mass is greater than that of a human being, its mass is not compared to the Earth's mass. In order to understand this problem we will use numerical data and the universal gravitation formula, to give validity to the explanation.

Force exerted by the Earth on a human being

$F=G*\frac{m_{1}*m_{2}}{r^2}$

Where:

G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]

m1 = mass of the person = 80 [kg]

m2 = mass of the earth 5.97*10^24[kg]

r = distance from the center of the earth to the surface or earth radius = 6371 *10^3 [m]

Now replacing we have

$F = 6.673*10^{-11} *\frac{80*5.97*10^{24}}{(6371*10^{3})^{2} } \\F = 785[N]$

Force exerted by a building on a human being

Where:

G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]

m1 = mass of the person = 80 [kg]

m2 = mass of the earth 300000 [ton] = 300 *10^6[kg]

r = distance from the building to the person = 2[m]

$F = 6.673*10^{-11}*\frac{80*300*10^6}{2^{2} } \\F= 0.4 [N]$

As we can see the force exerted by the Earth is 2000 times greater than that exerted by a building with the proposed data.

8 0

3 years ago

Force F = (-6.0 N)ihat + (2.0 N) jhatacts on a particle with position vector r =(4.0 m) ihat+ (4.0 m) jhat.

finlep [7]

Answer:

$\tau=(-32k)\ N-m$

$\theta=116.55^{\circ}$

Explanation:

Given that.

Force acting on the particle, $F=(-6i + 2.0j)\ N$

Position of the particle, $r=(4i+4j)\ m$

To find,

(a) Torque on the particle about the origin.

(b) The angle between the directions of r and F

Solution,

(a) Torque acting on the particle is a scalar quantity. It is given by the cross product of force and position. It is given by :

$\tau=F\times r$

$\tau=(-6i + 2.0j)\times (4i+4j)$

$\tau=\begin{pmatrix}0&0&-32\end{pmatrix}$

$\tau=(-32k)\ N-m$

So, the torque on the particle about the origin is (32 N-m).

(b) Magnitude of r, $|r|=\sqrt{4^2+4^2}=5.65\ m$

Magnitude of F, $|F|=\sqrt{(-6)^2+2^2}=6.324\ m$

Using dot product formula,

$F{\circ}\ r=|F|.|r|\ cos\theta$

$cos\theta=\dfrac{F{\circ} r}{|F|.|r|}$

$cos\theta=\dfrac{-24+8}{6.324\times 5.65}$

$\theta=116.55^{\circ}$

Therefore, this is the required solution.

6 0

3 years ago

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If ice has a density of.92g/cm, what is the volume of 1.8 kilograms of ice?

steposvetlana [31]

Given:
Density = .92 g / cm³
Volume = 1.8 kg
= (1.8 * 1000) grams
= 1800 grams
Now,
$Density (d) = \frac{Mass (m)}{volume(v)}$

$volume (v)= \frac{mass(m)}{density(d)}$

$volume (v)= \frac{1800}{.92}$

$volume (v)= 1956.521739$

$volume (v)= 1956.5~ cm^{3}$

So, the volume of 1.8 kg of ice is 1956.5 cm³

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3 years ago

Two astronauts push on a satellite. One pushes in the positive x direction with a force of 42 N, and the other pushes in the pos

inessss [21]

The vectors adition we can find the magnitude of the force applied by the other astronaut is 11.25 N in the y direction

Parameters given

Force of an astronaut Fₓ = 42 N

Angle θ = 15º

To find

Force another astronaut

The force is a vector magnitude for which the addition of vectors must be used, a very efficient method to perform this sum is to add the components of each vector and devise constructing the resulting vector using trigonometry and the Pythagorean theorem.

Let's use trigonometry to find the other force

tan θ = $\frac{F_y}{F_x}$

F_ y = Fₓ tan θ

let's calculate

F_y = 42 tan 15

F_y = 11.25 N

Using the summation of vectors we can find the magnitude of the force applied by the other astronaut is 11.25 N in the y direction

Learn more about vector addition here:

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3 years ago

Selena drew a diagram to show how current moves in a loop of wire that is placed between two magnets. At top left a piece of mag

OLEGan [10]

Answer:

B on Edge 2020

She can change the arrows so they show current traveling in opposite directions on the sides of the loop.

Explanation:

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3 years ago

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