All categories

3 years ago

When a high-mass star explodes, which of the following is left over in the center?

Physics

1 answer:

il63 [147K]3 years ago

8 0

I would say option D, it depends on the size of the star

You might be interested in

The absorption of longwave radiation in the atmosphere is popularly called ________.

Sliva [168]

I believe it is called the Greenhouse effect.

4 0

3 years ago

a student pushes a box with a total mass of 50kg. what is the net force on the box if it accelerates 1.5 m/s^2?

Zielflug [23.3K]

F=mass times acceleration so multiple 50 by 1.5 and u get 75

5 0

3 years ago

Magnesium hydroxide is a common ___?

omeli [17]

Answer:

Magnesium hydroxide is the inorganic compound with the chemical formula Mg(OH)2. Magnesium hydroxide is a common component of antacids, such as milk of magnesia, as well as laxatives

Explanation:

pls mark brainliest

6 0

3 years ago

Read 2 more answers

A mass weight of 120N is hung from two strings. what is the tension?

kramer

The weight should be shared between the two string equally. Therefore, tension in each string, T is;

T = 120 N/2 = 60 N

7 0

3 years ago

Read 2 more answers

A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h

Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

$v_{1}=\dfrac{3.0\times10^{3}}{3600}$

$v_{1}=0.833\ m/s$

The force F₁is constant acceleration is also a constant.

$F_{1}=ma_{1}$

We need to calculate the acceleration

Using formula of acceleration

$a_{1}=\dfrac{v}{t}$

$a_{1}=\dfrac{0.833}{10}$

$a_{1}=0.083\ m/s^2$

Similarly,

$F_{2}=ma_{2}$

For total force,

$F_{3}=F_{2}+F_{1}$

$ma_{3}=ma_{2}+ma_{1}$

The speed of second tugboat is

$v=\dfrac{11\times10^{3}}{3600}$

$v=3.05\ m/s$

We need to calculate total acceleration

$a_{3}=\dfrac{v}{t}$

$a_{3}=\dfrac{3.05}{10}$

$a_{3}=0.305\ m/s^2$

We need to calculate the acceleration a₂

$0.305=a_{2}+0.083$

$a_{2}=0.305-0.083$

$a_{2}=0.222\ m/s^2$

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

$\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}$

$\dfrac{F_{1}}{F_{2}}=3.7$

$F_{1}=3.7F_{2}$

Hence, The magnitude of F₁ is 3.7 times of F₂

3 0

3 years ago