All categories

3 years ago

This is for physics, i need help fast!!

Physics

1 answer:

jarptica [38.1K]3 years ago

7 0

Speed = distance / time
S= 40 000m / 5400s
S= 7.41m/s

You might be interested in

How does energy acquisition in the deep sea differ from energy acquisition near the ocean’s surface? a. Organisms in the deep se

Montano1993 [528]

that would be the energy from the core. because of the fact that ht eearths core is producing 45mill killawhats and giggawats

3 0

3 years ago

Read 2 more answers

Explain in terms of energy flow how a cold pack works on a sprained ankle

sergij07 [2.7K]

The <span>flow of how a cold pack works on a sprained ankle is based on the second law of thermodynamics which states that energy will flow from a higher to a lower temperature. So your body heat will flow to the cold pack in which you will feel the coldness of the pack.</span>

7 0

3 years ago

Help this Is easy for you

DochEvi [55]

Answer:

if its so easy why dont u do it .

Explanation:

6 0

3 years ago

The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$