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3 years ago

You make a straight line when an object is

Physics

2 answers:

swat323 years ago

8 0

Answer:

can you explain more or..,.

Explanation:

Nana76 [90]3 years ago

3 0

When you have two points, if you connect every point between those two points, you have a straight line.

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Which of the following is NOT a result of supernova explosions? The neutron core is completely destroyed. Any planets within a f

DanielleElmas [232]

Answer:

The neutron core is completely destroyed

Explanation:

A earth - supernova is an explosion resulting to the death of a star that occurs close enough to the earth but this does not completely destroy a star. Supernovae are the most violent explosions in the universe. But they do not explode like a bomb explodes, blowing away every bit of the original bomb. Rather, when a star explodes into a supernova, its core survives. The reason for this is that the explosion is caused by a gravitational rebound effect and not by a chemical reaction. Stars are so large that the gravitational forces holding them together are strong enough to keep the nuclear reactions from blowing them apart. It is the gravitational rebound that blows apart a star in a supernova.

4 0

2 years ago

The diagram shows four paths from point A to point B.

iVinArrow [24]

Path 2.

Displacement is the direction and magnitude of an object from its starting point, so path 2 is the direct route you would need to take to find direction and magnitude.

8 0

2 years ago

A car starts from rest and accelerates uniformly for a five seconds along a straight road. If speed obtained by the car is 72 km

Step2247 [10]

Answer:

50 meters

Explanation:

Let's start by converting to m/s. There are 3600 seconds in an hour and 1000 meters in a kilometer, meaning that 72km/h is 20m/s.

$v_f=v_o+at$

Since the car starts at rest, you can write the following equation:

$20=0+a(5) \\\\a=20\div 5=4 m/s^2$

Now that you have the acceleration, you can do this:

$d=v_o+\dfrac{1}{2}at^2$

Once again, there is no initial velocity:

$d=\dfrac{1}{2}(4)(5)^2=2 \cdot 25=50m$

Hope this helps!

8 0

3 years ago

Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What a

andrey2020 [161]

Explanation:

It is given that,

Force, $F=(-8\ N)i+(6\ N)j$

Position vector, $r=(3i+4j)\ m$

(a) The torque on the particle about the origin is given by :

$\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m$

(b) To find the angle between r and F use dot product formula as :

$F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}$

Hence, this is the required solution.

8 0

3 years ago

Read 2 more answers

We need to find the launch velocity of our new marble launcher. we know that it will launch a 25g marble to a distance of 73 cm,

White raven [17]

The launch velocity of the marble launcher is 34.65 m/s

Given that the launch velocity of marble launcher, launches a 25g marble to a distance of 73 cm (0.73 m) and the marble roll up to 6.2 meters before stopping. The launch height is 20 cm (0.2 m).

The time for landing can be calculated by the second equation of motion formula:

h = ut + $\frac{1}{2}$ g $t^{2}$