Which is the correct answer?

Based on the graph, which data point is most likely to have experimental

Len [333]

Answer:

B. 59 kg

Explanation:

From the graph you notice that a linear relation in indicated by the line joining the points such that the points on the line represent the data that show a correct relationship in the experiment.

This means that the point outside the line has an error .

This point is the value 59 kg that does not align with other values which are included in the graph.

8 0

2 years ago

A small fish is dropped by a pelican that is rising steadily at 0.50 m/s.

natima [27]

Answer:

(a) 24.025 m/s. downward.

(b) 31 m

Explanation:

From Newton's equation of motion,

(a)

v = u + gt ................... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, t = time.

Note: Let upward velocity be negative and downward be positive

Given: u = -0.5 m/s (upward), t = 2.5 s

Constant : g = 9.81 m/s²

Substitute into equation 1

v = 0.5+9.81(2.5)

v = -0.5+24.525

v = 24.025 m/s. downward.

(b) using

s₁ = ut + 1/2gt²......................... Equation 2

Where s₁ = distance at which the fish fall after being dropped by the pelican

Given: u = - 0.5 m/s, t = 2.5 s, g = 9.81 m/s²

Substitute into equation 2

s₁ = -0.5(2.5) + 1/2(9.81)(2.5)²

s₁ = -1.25+30.656

s₁ = 29.41 m

also,

s₂ = vt ................ Equation 3

Where s₂ = the distance by which the pelican rise during this time.

Given: v = 0.5 m/s, t= 2.5 s

s₂ = 0.5(2.5)

s₂ = 1.25 m.

Note: Distance between the pelican and fish = s₁ + s₂

Distance between the pelican and fish = 29.41+1.25

Distance between the pelican and fish = 30.66

Distance between the pelican and fish ≈ 31 m

3 0

3 years ago

A 1000 N object is pulled along a level surface with a horizontal force of 200 N. The object moves with a constant velocity of 2

krek1111 [17]

0.2 is the value of coefficient of friction (k)

F=kN

F=horizontal force

n=Normal Force

k=coefficient of friction

k=F/N

k=200/1000

k=0.2

The ratio of the normal force pushing two surfaces together to the frictional force preventing motion between them is known as the friction coefficient. Usually, the Greek letter mu is used to indicate it .N is the normal force, and F is the frictional force, hence F = N/N.

Due to the fact that both F and N are measured in units of force, the coefficient of friction has no dimensions (such as newtons or pounds). The coefficient of friction can have a variety of values for both static and dynamic friction. Static friction occurs when an object encounters friction that resists any applied force, keeping the object at rest until the static frictional force is released. In kinetic friction, the frictional force resists the motion of the object.

To know more about coefficient of friction visit brainly.com/question/136431

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7 0

1 year ago

provides some pertinent background for this problem. A pendulum is constructed from a thin, rigid, and uniform rod with a small

gavmur [86]

Answer:

the period of the physical pendulum is 0.498 s

Explanation:

Given the data in the question;

$T_{simple$ = 0.61 s

we know that, the relationship between T and angular frequency is;

T = 2π/ω ---------- let this be equation 1

Also, the angular frequency of physical pendulum is;

ω = √(mgL / $I$ ) ------ let this equation 2

where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and $I$ is moment of inertia of rod.

Now, moment of inertia of thin uniform rod D is;

$I$ = $\frac{1}{3}$ mD²

since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.

we substitute equation 2 into equation 1

we have;

T = 2π/ω OR T = 2π/√(mgL/ $I$ ) OR T = 2π√( $I$ /mgL)

so we can use $I$ = $\frac{1}{3}$ mD² for moment of inertia of the rod

Since center of gravity of the uniform rod lies at the center of rod

so that L = $\frac{1}{2}$ D.

now, substituting these equations, the period becomes;

T = 2π/√( $I$ /mgL) OR T = $2\pi \sqrt{\frac{\frac{1}{3}mD^2 }{mg(\frac{1}{2})D } }$ OR T = 2π√(2D/3g ) ----- equation 3

length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),

we have;

ω $_{simple$ = 2π/ $T_{simple$ OR ω $_{simple$ = √(g/D) OR ω $_{simple$ = 2π√( D/g )

so we simple solve for D/g and insert into equation 3

so we have;

T = √(2/3) × $T_{simple$

we substitute in value of $T_{simple$

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s

8 0

3 years ago

A 60 kilogram astronaut weighs 96 newtons on the surface of the moon. calculate the acceleration due to gravity on the moon.

34kurt

Given: Mass m = 60 Kg

Weight W = 96 N

Required: Acceleration due to gravity, g = ?

Formula: W = mg

g = W/g

g = 96 Kg.m/s²/60 Kg (note: this is the derive unit for Newton "N")

g = 1.6 m/s²

6 0

3 years ago

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