What's the name of a star that does 'not move'?

What type of reaction feels cold to the touch?

IRISSAK [1]

Answer:

Explanation:

An endothermic reaction absorbs heat, so it will be cold to touch.

8 0

3 years ago

April stands on a flat surface . If she has a mass of 72 what is the nirmal force acting on her

gayaneshka [121]

The normal force acting on April as she stands on a flat surface is 705.6 N.

Normal force of the girl

Fₙ = mg

where;

Fₙ is the normal force of the girl
m is mass of the girl
g is acceleration due to gravity

Fₙ = 72 x 9.8

Fₙ = 705.6 N

Thus, the normal force acting on April as she stands on a flat surface is 705.6 N.

Learn more about normal force here: brainly.com/question/14486416

#SPJ1

7 0

2 years ago

A car moved 50 km to the North. What is its displacement?

MariettaO [177]

Answer:

50 Km

Explanation:

If the car moved 50 km north and then 50 south its displacement would be 0. if it go in a direction like sw or ne or nw

4 0

3 years ago

Which of the following terms best describes why a skier sliding down a hill eventually comes to a stop?

natta225 [31]

Acceleration means you go faster
inertia means it has a tendency to do nothing or remain unchanged
gravity is what pulls things down to earth
<span>Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other.
so i would say friction</span>

6 0

3 years ago

Read 2 more answers

A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of

Lynna [10]

We have the relation

$\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}$

where $v_{A \mid B}$ denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

$v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}$

$v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}$

Let's assume the river flows South-to-North, so that

$\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath$

and let $-90^\circ < \theta < 90^\circ$ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

$\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)$

Then the velocity of the boat relative to the Earth is

$\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath$

The crossing is 153.0 m wide, so that for some time $t$ we have

$153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s$

which is minimized when $\theta=0^\circ$ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

$\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}$

The boat's position $\vec x$ at time $t$ is

$\vec x = \vec v_{B\mid E} t$

so that after 30.0 s, the boat's final position on the other side of the river is

$\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath$

and the boat would have traveled a total distance of

$\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}$

3 0

2 years ago