Earth is the only planet able to support what

PLS ANSWER FAST WILL GIVE BRAINLIEST! Use the image's labels to identify the location of each of the three subatomic particles.

DENIUS [597]

Answer:

Hey

A proton(s), and neutron(s).

B electron(s).

The nucleus (A) contains TWO particles, together they are called nucleons because they are in the nucleus. The two particles in the nucleus are the proton and neutron, they are held together by the nuclear strong force.

Outside of the nucleus, the electron takes up space around the nucleons. It is held in place by the electromagnetic force. It can only be a certain distance from the nucleus as you can see (the two circles around the nucleus). It can not be in between those two distances.

Wbob1314

5 0

2 years ago

An archer fires and arrow while standing atop a 5.15 m tall wall. The arrow is fired at an angle of 55 degrees and has a launch

il63 [147K]

Answer:

Explanation:

Given

height of wall=5.15 m

angle of launch $(\theta )=55^{\circ}$

Launch velocity(u)=52.4 m/s

Time of flight will be sum of time of flight of projectile+time to cover 5.15 m

Time of flight of arrow $=\frac{2usin\theta }{g}$

$t=\frac{2\times 52.4\times sin55}{9.81}=8.76 s$

Now time require to cover 5.15 m

Here at the time of zero vertical displacement of arrow i.e. when arrow is at the same height as of building then its vertical velocity will change its sign compared to initial vertical velocity.

$v_y=-52.4sin55$ at zero vertical displacement

Thus time required will be $t_2$

$5.15=52.4sin55\times t+\frac{gt^2}{2}$

$4.9t^2+42.92t-5.15=0$

$t=0.118 s i.e. t_2=0.118 s$

total time = $t_1+t_2=8.76+0.118=8.878 s$

(b)Horizontal distance=Range of arrow(R_1) + horizontal distance in 0.118 s $(R_2)$

$R_1=\frac{u^2sin2\theta }{g}=\frac{52.4^2\times sin110}{9.8}=263.28 m$

$R_2=52.4cos55 \times 0.118=3.546 m$

$R=R_1+R_2$ =263.28+3.546=266.82 m

8 0

3 years ago

Consult Multiple Concept Example 10 in preparation for this problem. Traveling at a speed of 18.2 m/s, the driver of an automobi

Anettt [7]

Answer:

The speed of the automobile after 1.43s is 10 $\frac{m}{s}$

Explanation:

$a= \frac{-f}{m}= \frac{-u_{k}*m*g}{m}$

$a= -u_{k}*g=- 0.590* 9.8 \frac{m}{s^{2} }= -5.782 \frac{m}{s^{2} }$

$V_{f} = V_{i} + a*t$

$V_{f} = 18.2 \frac{m}{s} - (5.782 \frac{m}{s^{2} }* 1.43 s)$

$V_{f} = 9.93174 \frac{m}{s}$

$V_{f}$ ≅ 10 $\frac{m}{s}$

7 0

3 years ago

The interference of two sound waves of slightly different frequencies produces

Leviafan [203]

Beats.

When two sound waves of different frequency approach your ear, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud - a phenomenon which is called "beating" or producing beats. The beat frequency is equal to the absolute value of the difference in frequency of the two waves.

4 0

3 years ago

Read 2 more answers

A stone that is thrown vertically upwards was having a velocity of 15m/s after reaches 2/3 of its maximum height. What is the ma

attashe74 [19]

The correct answer is option B. The maximum height that can be reached by the stone is determined as 11.5 m.

<h3>Maximum height attained by the stone </h3>

The maximum height attained by the stone when it is a 2/3 of its total height is calculated as follows;

v² = u² - 2gh

where;

v is final velocity at maximum height, v = 0
u is initial velocity
g is acceleration due to gravity

0 = u² - 2gh

2gh = u²

h = u²/2g

h = (15²)/(2 x 9.8)

h = 11.48 m

h = 11.5 m

Thus, the maximum height that can be reached by the stone is determined as 11.5 m

Learn more about maximum height here: brainly.com/question/12446886

#SPJ1

4 0

2 years ago