All categories

3 years ago

PLZZZ HELP 100 POINTS WILL MARK BRAINELST

Physics

2 answers:

Kitty [74]3 years ago

3 0

Explanation:

Crust...molten

a. Oceanic, iron

b. Continental, silicates

c. less

3. Mantle, Denser

a. Lithosphere

b. Asthenosphere

4. Core

a. elements, rocks

b. liquid, magnetic

(I guess the liquid should come after the is)

Couldn't answer all but wanted to help

shtirl [24]3 years ago

3 0

Answer:

<h2><em><u>Crust...molten </u></em></h2><h2><em><u> a. Oceanic, iron </u></em></h2><h2><em><u> b. Continental, silicates </u></em></h2><h2><em><u> c. less </u></em></h2><h2><em><u> 3. Mantle, Denser </u></em></h2><h2><em><u> a. Lithosphere </u></em></h2><h2><em><u> b. Asthenosphere </u></em></h2><h2><em><u> 4. Core </u></em></h2><h2><em><u> a. elements, rocks </u></em></h2><h2><em><u> b. liquid, magnetic </u></em></h2><h2><em><u> (I guess the liquid should come after the is) </u></em></h2><h2><em><u> </u></em></h2>

You might be interested in

a 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.88 m/s left. afterwards, the 282 kg car

Vaselesa [24]

The momentum of the 155 kg car afterwards is 469.7 kg m/s to the right

Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum of the system is conserved before and after the collision, so we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 282 kg$ is the mass of the bumper car

$u_1 = 3.50 m/s$ is the initial velocity of the bumper car (we take the right as positive direction)

$v_1 = 0.800 m/s$ is the final velocity of the bumper car

$m_2 = 155 kg$ is the mass of the second bumper car

$u_2 = -1.88 m/s$ is the initial velocity of the second car (moving to the left)

$v_2$ is the final velocity of the second car

Solving for $v_2$ ,

$v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(282)(3.50)+(155)(-1.88)-(282)(0.800)}{155}=3.03 m/s$

where the positive sign means the direction is to the right.

And now we can find the momentum of the 155 kg afterwards, which is

$p_2 = m_2 v_2 = (155)(3.03)=469.7 kg m/s$ (to the right)

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

7 0

3 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago

A dart is loaded into a spring-loaded toy dart gun by pushing the spring in by a  distance d. For the next loading?the spring is

Mrac [35]

Answer:

$W'=\dfrac{W}{9}$

Explanation:

The potential energy of the spring or the work done by the spring is given by :

$W=\dfrac{1}{2}kd^2$ ............(1)

k is the spring constant

d is the compression

When the spring is compressed a distance d' = d/3, let W' is the work is required to load the second dart. Then the work done is given by :

$W'=\dfrac{1}{2}kd'^2$

$W'=\dfrac{1}{2}k(d/3)^2$ .............(2)

Dividing equation (1) and (2) :

$\dfrac{W}{W'}=\dfrac{1/2kd^2}{1/2k(d/3)^2}$

$\dfrac{W}{W'}=9$

$W'=\dfrac{W}{9}$

So, the work required to load the second dart compared to that required to load the first is one-Ninth as much. Therefore, the correct option is (E).

7 0

3 years ago

Ok, so this question is probably really easy but I can't really be bothered to answer it, terrible I know, but I thank all usefu

Dvinal [7]

Without a bulb energy cant go through and it would be an open circuit blocking the energy from coming out.

3 0

3 years ago

A wire carrying a 2-A current is placed at an angle of 60° with the respect to a magnetic field of strength 0.2 T. If the length

Anvisha [2.4K]

Answer:

0.208 N

Explanation:

Parameters given:

Current, I = 2A

Angle, A = 60°

Magnetic field strength, B = 0.2 T

Length, L = 0.6 m

Magnetic force is given as:

F = I * L * B * sinA

F = 2 * 0.6 * 0.2 * sin60

F = 0.24 * sin60.

F = 0.208 N

8 0

3 years ago

Read 2 more answers