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Vikki [24]
2 years ago
6

An eagle accelerates at 22.35 m/s2 with a force of 120 N. Disregarding other forces acting on the eagle such as gravity and air

resistance, calculate the mass of the eagle. Round the answer to the nearest tenth of a kg.
The mass of the eagle is
kg.
Physics
2 answers:
Troyanec [42]2 years ago
6 0

Answer:

Mass of the eagle, m = 5.36 kg

Explanation:

It is given that,

Acceleration of the eagle, a = 22.35 m/s²

Force acting on eagle, F = 120 N

We need to find the mass of the eagle. It is given by using the second law of motion as :

F = m a

m=\dfrac{F}{a}

m=\dfrac{120\ N}{22.35\ m/s^2}

m = 5.36 kg

So, the mass of the eagle is 5.36 kg. Hence, this is the required solution.

andrew11 [14]2 years ago
3 0
5.4 kg is the answer on edgenuity
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A construction worker accidentally drops a brick from a high scaffold. a. What is the brick's velocity after 4.0 s? b. How far d
AlekseyPX

Answer:

A. 39.2 m/s

B. 78.4 m

Explanation:

Data obtained from the question include:

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

A. Determination of the brick's velocity.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = gt

v = 4 × 9.8

v = 39.2 m/s

Thus, the brick's velocity after 4 s is 39.2 m/s

B. Determination of how far the brick fall in 4 s.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

h = ½gt²

h = ½ × 9.8 × 4²

h = 4.9 × 16

h = 78.4 m

Thus, the brick fall 78.4 m during the time.

5 0
2 years ago
A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i
victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

F= \frac{kqQ}{r^2}

Where,

k = 8.99 * 10^9 N m^2 / C^2.

q = charges of the objects

r= distance/radius

Our values are previously given, so

q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59

Replacing,

F=\frac{kqQ}{r^2}

F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}

F= 4.8423N

The force acting on the block are given by,

F-mgsin\theta = ma

a = \frac{F-mgsin\theta}{m}

a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}a = 10.31m/s^2

Therefore the box is accelerated upward.

3 0
2 years ago
Match each projection to its description.
babymother [125]
A goes with 2 and B goes with 1.
6 0
2 years ago
Read 2 more answers
Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculat
kodGreya [7K]

Answer:

1.6675×10^-16N

Explanation:

The force of gravity that the space shuttle experiences is expressed as;

g = GM/r²

G is the gravitational constant

M is the mass = 1.0 x 10^5 kg

r is the altitude = 200km = 200,000m

Substitute into the formula

g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²

g = 6.67×10^-6/4×10^10

g = 1.6675×10^{-6-10}

g = 1.6675×10^-16N

Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N

7 0
2 years ago
In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal
SpyIntel [72]

Answer:

A) F_g = 26284.48 N

B) v = 7404.18 m/s

C) E = 19.19 × 10^(10) J

Explanation:

We are given;

Mass of satellite; m = 3500 kg

Mass of the earth; M = 6 x 10²⁴ Kg

Earth circular orbit radius; R = 7.3 x 10⁶ m

A) Formula for the gravitational force is;

F_g = GmM/r²

Where G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Plugging in the relevant values, we have;

F_g = (6.67 × 10^(-11) × 3500 × 6 x 10²⁴)/(7.3 x 10⁶)²

F_g = 26284.48 N

B) From the momentum principle, we have that the gravitational force is equal to the centripetal force.

Thus;

GmM/r² = mv²/r

Making v th subject, we have;

v = √(GM/r)

Plugging in the relevant values;

v = √(6.67 × 10^(-11) × 6 x 10²⁴)/(7.3 x 10⁶))

v = 7404.18 m/s

C) From the energy principle, the minimum amount of work is given by;

E = (GmM/r) - ½mv²

Plugging in the relevant values;

E = [(6.67 × 10^(-11) × 3500 × 6 × 10²⁴)/(7.3 x 10⁶)] - (½ × 3500 × 7404.18)

E = 19.19 × 10^(10) J

5 0
2 years ago
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