Please help need answer asp

A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the si

solmaris [256]

Answer:

4.384 * 10^13

Explanation:

Given the expression :

[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]

Applying the laws of indices

[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]

13.2733 * 10^19 ÷ 3.0276 * 10^6

(13.2733 / 3.0276) * 10^(19 - 6)

4.3840996 * 10^13

= 4.384 * 10^13

6 0

2 years ago

1.) I’m bored, so I decide to play catch by myself. I throw a 1.5kg ball in the air at 25 m/s. How long do I have to wait to cat

navik [9.2K]

Answer:

technically yes

Explanation:

with a gun depending on how fast it shoots so when you fire at something you shoot in front of it a little bit so you hit it but a plane that fast you shoot like 100 feet infront of it...

4 0

3 years ago

Read 2 more answers

An object glides on a horizontal tabletop with a coefficient of kinetic friction of 0.5. If its initial velocity is 4.3 m/s, how

Shkiper50 [21]

Answer:

Time, t = 0.87 seconds

Explanation:

Given that,

Initial velocity of the object, u = 4.3 m/s

The coefficient of kinetic friction between horizontal tabletop and the object is 0.5

We need to find the time taken by the object for the object to come to rest i.e. final velocity will be 0.

Using first equation of motion to find it as :

$v=u+at$

a is the acceleration, here, $a=\mu g$

$0=u+\mu gt$

$t=\dfrac{u}{\mu g}\\\\t=\dfrac{4.3}{0.5\times 9.8}\\\\t=0.87\ s$

So, the time taken by the object to come at rest is 0.87 seconds. Hence, this is the required solution.

8 0

3 years ago

In this graph, what is the displacement of the particle in the last two seconds?

nikklg [1K]

In this graph, what is the displacement of the particle in the last two seconds?of the particle in the last two seconds?

<span>0.2 meters
2 meters
4 meters
6 meters</span>
In this graph, the displacement of the particle in the last two seconds is 2 meters.

7 0

3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).