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3 years ago

Two bulldozers are pushing on a large tree trunk as shown. (last photo)

Physics

1 answer:

andrey2020 [161]3 years ago

8 0

The tree will fall in the south eastern side of the area due to two bulldozer.

<h3>Direction of tree falling</h3>

The tree falls in the sloppy direction due to both bulldozer which apply force from two directions. One bulldozer apply force on the tree from western side whereas the other bulldozer apply force on the tree from north side so that's why we can say that the tree will fall in the south eastern side of the area.

Learn more about force here: brainly.com/question/12970081

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HELP PLEASE!!!

irakobra [83]

Answer:

10kN

Explanation:

Given data

m1= 50kg

u1= 3m/s

m2= 100kg

u2= 6m/s

v1= 2m/s

time= 0.04s

let us find the final velocity of Bruce v1

from the conservation of linear momentum

m1u1+m2u2=m1v1+m2v2

substitute

50*3+100*6= 50*v1+100*2

150+600=50v1+200

750-200=50v1

550= 50v1

divide both sides by 50

v1= 550/50

v1=11 m/s

From

F= mΔv/t

for Bruce

F=50*(11-3)/0.04

F=50*8/0.04

F=400/0.04

F=10000

F=10kN

for Max

F=100*(6-2)/0.04

F=100*4/0.04

F=400/0.04

F=10000

F=10kN

6 0

3 years ago

How does the total momentum of the docked vehicle and station compare to the momentum of each object before the

Xelga [282]

The total momentum before the docking maneuver is and after the docking maneuver is

Explanation:

Linear momentum (generally just called momentum) is defined as mass in motion and is given by the following equation:

Where is the mass of the object and its velocity.

According to the conservation of momentum law:

"If two objects or bodies are in a closed system and both collide, the total momentum of these two objects before the collision will be the same as the total momentum of these same two objects after the collision ".

This means, that although the momentum of each object may change after the collision, the total momentum of the system does not change.

Now, the docking of a space vehicle with the space station is an inelastic collision, which means both objects remain together after the collision.

Hence, the initial momentum is:

Where:

is the mass of the vehicle

is the velocity of th vehicle

is the mass of the space station

is the velocity of the space station

And the final momentum is:

Where:

is the velocity of the vehicle and space station docked

5 0

4 years ago

How would seasons on earth be different if the earth was not tilted on its axis

MakcuM [25]

If the Earth's axis were 'straight' ... pefectly perpendicular to the ecliptic
plane ... then:

-- Day and night would be the same length ... every day of the year,
everywhere on Earth !

-- There wouldn't be any seasons, anywhere. There might still be some
'weather' ... cloudy days, sunny days, occasional rain, wind etc. But
there would be no average change during the year. No hot months or
cold months. In any one place, the weather would always be generally
the same, every day, all year. Everywhere all around the equator would be
generally the hottest on Earth, and the local climates would generally get
cooler as you moved away from the equator and toward the poles.

5 0

4 years ago

At high noon, the sun delivers 1150 w to each square meter of a blacktop road. if the hot asphalt loses energy only by radiation

Pie

Stephen`s Law:
P = (Sigma) · A · e · T^4
P in = P out
e = 1 for blacktop;
1150 W = (Sigma) · T^4
(Sigma) = 5.669 · 10 ^(-8) W/m²K^4
T^4 = 1150 : ( 5.669 · 10^(-8) )
T^4 = 202.875 · 10^8
$T = \sqrt[4]{202.857 * 10 ^{8} }$
T = 3.774 · 10² = 377.4 K
Answer: Equilibrium temperature is 377.4 K.

3 0

3 years ago

A 77.0 kg woman slides down a 42.6 m long waterslide inclined at 42.3. at the bottom, she is moving 20.3 m/s.how much work did f

svlad2 [7]

Answer:

W = 21409.2 J

Explanation:

Given,

The mass of the woman, m = 77 Kg

The length of the water slide, S = 42.6 m

The inclination of the water slide, ∅ = 42.3

The constant velocity of the women sliding, 20.3 m/s

The kinetic friction of the sliding force is given by the formula

Fₓ = μₓ η

Where,

μₓ - coefficient of kinetic friction

η - normal force acting on the body

Since the water slide is inclined at an angle and the person is sliding with constant velocity. The coefficient of friction becomes,

μₓ = tan∅

And, η = mg cos∅

Therefore, the kinetic friction force becomes

Fₓ = tan∅ mg cos∅

Substituting the given values in the above equation

Fₓ = 0.9 x 77 x 9.8 x 0,74

= 502.56 N

The work done by the kinetic friction on the person

W = Fₓ · S

= 502.56 N x 42.6 m

= 21409.2 J

Hence, the work done by the friction on the woman is, W = 21409.2 J

8 0

3 years ago