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satela [25.4K]
2 years ago
6

3) A bird flies toward a tree limb at a 45-degree angle to the ground along a path that is 50 m long landing on the limb. Determ

ine, in m, how high above the ground the bird is perched.
Physics
1 answer:
marin [14]2 years ago
6 0

Answer:

35 m

Explanation:

Given :

The distance of the path from the ground to the tree limb = 50 m

The angle between the path of flight of the bird towards the tree limb and the ground = 45 degrees

Therefore we can determine the height above which the bird perched above the ground by using the rules of the trigonometric ratios as;

We know that ,

$\sin 45^\circ =\frac{\text{perpendicular}}{\text{hypothenus}}$

$\sin 45^\circ =\frac{\text{BC}}{\text{AC}}$

$0.7  =\frac{\text{BC}}{\text{50}}$

\text{BC }= 0.7 \times 50

       = 35

Therefore, the bird perched on the tree limb at a height of 35 m.

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Gravitational force depends on the mass of an object.
vampirchik [111]

Answer:

sun, earth, capitol building, human, atom

Explanation:

Gravity is directly proportional to the size of an object. Therefore, the object with the most mass will have the greatest force of gravity. We know this from the equation. F = G\frac{m_1m_2}{r^2}

4 0
2 years ago
A window in a skyscraper has a surface area of 3.50 m^2. Wind rushes by the outside of the window at 17.4 m/s, while inside the
Mariana [72]

The difference in the pressure between the inside and outside will be 369.36 N/m²

<h3>What is pressure?</h3>

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.

It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.

The given data in the problem is;

dP is the change in the presure=?

Using Bernoulli's Theorem;

\rm  \rho\frac{V^2_{12}}{2} +P_1= \rho \frac{V^2_{22}}{2} +P_2 \\\\\ P_2-P_1=\rho \frac{v_2^2-v_1^2}{2} \\\\  P_2-P_1= 1.21 \times \frac{17.4^2-0}{2} \\\\  \triangle p=369.36 \ N/m^2

Hence, the difference in the pressure between the inside and outside will be 369.36 N/m²

To learn more about the pressure refer to the link;

brainly.com/question/356585

#SPJ1

3 0
2 years ago
In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m
ikadub [295]

(a) -39.4^{\circ}

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

p_y =0

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2

where

m is the mass of each ball

v_1= 4.60 m/s is the velocity of the cue ball after the collision

v_2 = 3.40 m/s is the velocity of the second ball after the collision

\phi_1=28.0^{\circ} is the angle of the cue ball with the x-axis

\phi_2 is the angle of the second ball

Solving for \phi_2, we find the angle between the direction of motion of the second ball and the original direction of motion:

sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m

The initial total momentum along the x-direction as

p_x = m u

where

m is the mass of the cue ball

u is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

mu = 6.69 m\\u = 6.69 m/s

7 0
3 years ago
49. A block is pushed across a horizontal surface with a
nata0808 [166]

Answer:

(a) 37.5 kg

(b) 4

Explanation:

Force, F = 150 N

kinetic friction coefficient = 0.15

(a) acceleration, a = 2.53 m/s^2

According to the newton's second law

Net force = mass x acceleration

F - friction force = m a

150 - 0.15 x m g = m a

150 = m (2.53 + 0.15 x 9.8)

m = 37.5 kg

(b) As the block moves with the constant speed so the applied force becomes the friction force.

F = \mu m g \\\\150 = \mu\times 37.5\\\\\mu = 4

8 0
3 years ago
A body, with a volume of 2 m3, weighs 40 kN. Determine its weight when
lana66690 [7]

Answer:

8.8 kN

Explanation:

V = 2 m³, W = 40 kN, SG = 1.59

Bouyant force N = 1.59 * 1000 kg/m³ * 9.81 N/kg * 2 m³ = 31.2 kN

So the weight becomes 40 - 31.2 = 8.8 kN

3 0
2 years ago
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