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baherus [9]
3 years ago
13

An eagle is flying horizontally 16.4 meters above a lake at a speed of 9.3 m/s, carrying a small pumpkin in its talons. The pump

kin slips free. How far horizontally will the pumpkin travel after it slips from the eagle until it hits the ground?
Physics
1 answer:
Dima020 [189]3 years ago
4 0

Answer:

The horizontal distance the pumpkin will travel after it slips from the eagle is 17.02 m

Explanation:

Given;

height above the ground, h = 16.4 m

speed of the eagle, v = 9.3 m/s

The time it will take the pumpkin to fall at the given height is calculated as;

t = \sqrt{\frac{2h}{g} }\\\\t =  \sqrt{\frac{2*16.4}{9.8} }\\\\t = 1.83 \ s

The horizontal distance traveled at this time is given by;

x = vt

x = (9.3)(1.83)

x = 17.02 m

Therefore, the horizontal distance the pumpkin will travel after it slips from the eagle is 17.02 m

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How will the equation of morion for an object moving with an ununifor motion change????? please help me !!!!!!
babymother [125]

Answer:

The equation of motion is derived based on the Newton’s laws of motion. And it changes accordingly when an object changes with uniform velocity.

Given is that object moves with uniform velocity, that is no change in velocity so there will no acceleration.

As we know    

Here, u = v (due to uniform velocity)

.

1st equation of motion is, v = u + at

2nd equation of motion,  

3rd equation of motion,  

.

Explanation:

4 0
3 years ago
Ball A (mass of 0.5 kg) going +2.0 m/s collides with a stationary ball B (mass of 0.4 kg).
ki77a [65]

The correct answer is 1.2 m/s

: mv+mv=mv+mv

(0.5kg)(2m/s)+(0.4kg)(0m/s)=(0.5kg)v+(0.4kg)(1m/s)

= 1kg*m/s=(0.5kg)v+0.4kg*m/s

=1kg*m/s-0.4kg*m/s=(0.5kg)v

=0.6kg*m/s=(0.5kg)v

to solve for v we divide both side by 0.5kg

v=1.2m/s

6 0
3 years ago
How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight
Natalka [10]
By using Ohm's law, we can find what should be the resistance of the wire, R:
R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega

Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
r=35 mm=0.35 \cdot 10^{-3} m
So the area is
A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2

And by using the resistivity  of the Aluminum, \rho=2.65 \cdot 10^{-8} \Omega m, we can use the relationship between resistance R and resistivity:
R= \frac{\rho L}{A}
to find L, the length of the wire:
L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m
4 0
3 years ago
Collapse question part Part 4 (d) What is the unit vector in the direction of the spacecraft's velocity? (Express your answer in
Maksim231197 [3]

Answer:

unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]

Explanation:

Given:

                            v =  (-23.2, -104.4, 46.4) m/s

Above expression describes spacecraft's velocity vector v.

Find:

Find unit vector in the direction of spacecraft velocity v.

Solution:

Step 1: Compute magnitude of velocity vector.

                            mag (v) = sqrt ( 23.2^2 + 104.4^2 + 46.4^2)

                            mag (v) = 116.58 m/s

Step 2: Compute unit vector unit (v)

                            unit (v) = vec (v) / mag (v)

                            unit (v) = [ -23.2 i -104.4 j + 46.4 k ] / 116.58

                            unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]

7 0
3 years ago
The specific gravity of mantle rock is about 3.3. True False
Alla [95]
True is the correct answer
5 0
3 years ago
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