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3 years ago

NEED ANSWER NOW PLZ WILL MARK BRAINLIEST

Physics

2 answers:

PilotLPTM [1.2K]3 years ago

8 0

I'm like 89% sure that the answer is C.

Kay [80]3 years ago

8 0

it is the process by which important molecules enter the body cells and waste products are removed from cells. is why it is important because your body kills off the stuff that's not supposed to be in it and makes sure all the good cells are where they're supposed to be

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Two hockey pucks with mass 0.1 kg slide across the ice and collide. Before

RideAnS [48]

The velocity of pluck 1 is 12 m/s west.

<h3>What is the conservation of momentum?</h3>

The principle of the conservation of the linear momentum states that momentum before collision is equal to momentum after collision.

Now given that;

m1u1 + m2u2 = m1v1 + m2v2

(0.1 * 15) - (0.1 * 12) = 0.1* v + (0.1 * 15)

1.5 - 1.2 = 0.1v + 1.5

0.3 - 1.5 = 0.1v

v = -1.2/0.1

v = - 12 m/s

Hence, the velocity of pluck 1 is 12 m/s west.

Learn more about linear momentum:brainly.com/question/27988315

#SPJ1

7 0

1 year ago

The rotational speed of earth is similar to?

m_a_m_a [10]

Actually, the speed of the earth is the same everywhere, taking the angular speed as the valid measure of the speed

8 0

3 years ago

Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E

Kay [80]

Answer:

The true course: $40.29^\circ$ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

$V_w$ = velocity of wind = $30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h$

$V_p$ = velocity of plane in still air = $100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h$

Assume:

$V_r$ = resultant velocity of the plane

$\theta$ = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

$\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h$

Let us find the direction of this resultant velocity with respect to east direction:

$\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ$

This means the the true course of the plane is in the direction of $40.29^\circ$ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

$\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h$

Hence, the ground speed of the plane is 96.68 km/h.

5 0

3 years ago

Vector A has a magnitude of 50 units and points in the positive x direction. A second vector, B , has a magnitude of 120 units a

Alex Ar [27]

A) Vector A

The x-component of a vector can be found by using the formula

$v_x = v cos \theta$

where

v is the magnitude of the vector

$\theta$ is the angle between the x-axis and the direction of the vector

- Vector A has a magnitude of 50 units along the positive x-direction, so $\theta_A = 0^{\circ}$ . So its x-component is

$A_x = A cos \theta_A = (50) cos 0^{\circ}=50$

- Vector B has a magnitude of 120 units and the direction is $\theta_B = -70^{\circ}$ (negative since it is below the x-axis), so the x-component is

$B_x = B cos \theta_B = (120) cos (-70^{\circ})=41$

So, vector A has the greater x component.

B) Vector B

Instead, the y-component of a vector can be found by using the formula

$v_y = v sin \theta$

Here we have

- Vector B has a magnitude of 50 units along the positive x-direction, so $\theta_A = 0^{\circ}$ . So its y-component is

$A_y = A sin \theta_A = (50) sin 0^{\circ}=0$

- Vector B has a magnitude of 120 units and the direction is $\theta_B = -70^{\circ}$ , so the y-component is

$B_y = B sin \theta_B = (120) sin (-70^{\circ})=-112.7$

where the negative sign means the direction is along negative y:

So, vector B has the greater y component.

8 0

3 years ago

A small child has a wagon with a mass of 10 kilograms. The child pulls on the wagon with a force of

Sholpan [36]

$F=ma$
where:
F - force
m - mass
a - acceleration

We transform this formula to get a:
$a= \frac{F}{m}$
$a=\frac{2}{10}\frac{N}{kg}=0.2\frac{m}{s^{2}}$

4 0

2 years ago

Read 2 more answers