Answer:
the size of the shadow will be smaller due to smaller hands
Density = (mass) / (volume)
4,000 kg/m³ = (mass) / (0.09 m³)
Multiply each side
by 0.09 m³ : (4,000 kg/m³) x (0.09 m³) = mass
mass = 360 kg .
Force of gravity = (mass) x (acceleration of gravity)
= (360 kg) x (9.8 m/s²)
= (360 x 9.8) kg-m/s²
= 3,528 newtons .
That's the force of gravity on this block, and it doesn't matter
what else is around it. It could be in a box on the shelf or at
the bottom of a swimming pool . . . it's weight is 3,528 newtons
(about 793.7 pounds).
Now, it won't seem that heavy when it's in the water, because
there's another force acting on it in the upward direction, against
gravity. That's the buoyant force due to the displaced water.
The block is displacing 0.09 m³ of water. Water has 1,000 kg of
mass in a m³, so the block displaces 90 kg of water. The weight
of that water is (90) x (9.8) = 882 newtons (about 198.4 pounds),
and that force tries to hold the block up, against gravity.
So while it's in the water, the block seems to weigh
(3,528 - 882) = 2,646 newtons (about 595.2 pounds) .
But again ... it's not correct to call that the "force of gravity acting
on the block in water". The force of gravity doesn't change, but
there's another force, working against gravity, in the water.
Assuming that the object starts at rest, we know the following values:
distance = 25m
acceleration = 9.81m/s^2 [down]
initial velocity = 0m/s
we want to find final velocity and we don't know the time it took, so we will use the kinematics equation without time in it:
Velocity final^2 = velocity initial^2 + 2 × acceleration × distance
Filling everythint in, we have:
Vf^2 = 0^2 + (2)(-9.81)(-25)
The reason why the values are negative is because they are going in the negative direction
Vf^2 = 490.5
Take the square root of that
Final velocity = 22.15m/s which is answer c
Answer:
The answer of the part (a) is v2 = 7.09 m/s
and the answer of the part (b) is vA1 = 5.25 m/s
Explanation:
Explanation of the both parts of answer is in the following attachments
The force which has the greatest effect on causing this object to slow while it remains in contact with the ramp is: B. a frictional force.
<h3>What is a force?</h3>
A force can be defined as a push or pull of an object or physical body, which typically results in a change of motion (acceleration), especially due to the interaction of the object with another.
<h3>The types of force.</h3>
In Science, there are different types of force and these include the following:
<h3>What is a
frictional force?</h3>
Friction force can be defined as a type of force that resists and slows the relative motion of two physical objects when there surfaces come in contact. This ultimately implies that, a frictional force prevents two surfaces from easily sliding over or slipping across one another.
In this context, we can infer and logically deduce that the force which has the greatest effect on causing this object to slow while it remains in contact with the ramp is a frictional force.
Read more on frictional force here: brainly.com/question/25253774
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Complete Question:
Brandon pushes an object on a ramp as shown in the diagram.
While Brandon pushes the object and it remains in contact the ramp, which force has the greatest effect on causing it to slow?
A. the applied force
B. a frictional force
C. the force due to gravity
D. a force of air resistance