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2 years ago

A snowball starting at rest rolls down a hill and reaches 5 m/s. If the hill is

Physics

1 answer:

Lostsunrise [7]2 years ago

4 0

Answer:

The acceleration of the snowball is 0.3125

Explanation:

The initial speed of the snowball up the hill, u = 0

The speed the snowball reaches, v = 5 m/s

The length of the hill, s = 40 m

The equation of motion of the snowball given the above parameters is therefore;

v² = u² + 2·a·s

Where;

a = The acceleration of the snowball

Plugging in the values, we have;

5² = 0² + 2 × a × 40

∴ 2 × 40 × a = 5² = 25

80 × a = 25

a = 25/80 = 5/16

a = The acceleration of the snowball = 5/16 m/s².

The acceleration of the snowball = 5/16 m/s² = 0.3125 m/s² .

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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

2 years ago

Looking at the bottom standing wave in the picture, how many complete waves are present?

Shtirlitz [24]

Answer:

the last one

Explanation:

4 0

1 year ago

Orbitals that are of equal energy may be referred to as.

ivanzaharov [21]

Answer:

degenerate

Explanation:

5 0

1 year ago

Two crates, one with mass 5.4 kg and the other with mass 8.2 kg, connected by a light rope. The coefficient of kinetic friction

Gnom [1K]

Answer:

R= 2.5 :ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks

Explanation:

We apply Newton's second law:

∑F=m*a

velocity is constant ,then , a=0

Nomenclature

W: weight

m: mass

N : normal force

Ff: Friction force

μk: coefficient of kinetic friction

T: tension force in the rope

F: applied horizontal force

g: acceleration due to gravity.

Force Calculation

W₁=m₁*g=5.4 kg *9.8m/s²=52.92 N

W₂=m₂*g=8.2 kg *9.8m/s²= 80.36N

∑Fy=0

N₁-W₁=0 , N₁=W₁ = 52.92 N

N₂-W₂=0, N₂=W₂=80.36N

Ff₁= μk* N₁=0.4*52.92 N = 21.16N

Ff₂= μk* N₂=0.4*80.36N = 32.14N

Look at the attached graphic

Free-Body diagram m₁=5.4 kg

∑Fx=0

T- Ff₁=0 , T= Ff₁ , T= 21.16N

Free-Body diagram m₂=8.2 kg

∑Fx=0

F-T- Ff₂=0 , F=T+Ff₂= 21.16N+32.14N=53.3N

Ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks (R)

R= F/T= 53.3N/21.16N = 2.5

3 0

2 years ago

She leaves her campsite and runs north for 10 miles turns east and runs 8 miles then turns south and runs 2 miles.whats her dist

Mama L [17]

The distance is 20 miles

Explanation:

Definitions:

Distance is the total length of the path covered by an object during its motion, regardless of the direction of its motion
Displacement is the shortest distance between the initial and final position of the motion, so it also takes into account the direction of motion

In this problem, we want to know the distance covered by the woman. The motion consists of the following parts:

$d_1 = 10 miles$ north