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2 years ago

Please help me with this very easy question I just don’t get it.

Physics

1 answer:

diamong [38]2 years ago

5 0

Answer:

150m

Explanation:

The relation of speed/time and distance/time is a derivative/integral one, as in speed is the derivative of distance (the faster you go, the faster the distance changes, duh!).

So we need to compute the integral of speed over time from 0.0s to 5.0s.

The easiest way here is to compute the area under the line (it's going to be faster than computing the acceleration and using a formula of distance based on acceleration).

The area under the line is a trapezoid with "height" 5s, and the bases 10m/s and 50m/s. Using the trapezoid area formula of h*(a + b)/2

distance = 5s * (10m/s + 50m/s) / 2 = 5s * 60m/s / 2 = 5s * 30m/s = 150m

Alternatively, we can use the acceleration formula:

a = (50m/s - 10m/s)/5s = 40m/s / 5s = 8m/s^2

distance = v0 * t + a * t^2 / 2 = 10m/s * 5s + 8m/s^2 * (5s)^2 / 2 = 50m + 8m * 25 / 2 = 50m + 100m = 150m.

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A boy in a wheelchair (total mass 54.5 kg) has speed 1.40 m/s at the crest of a slope 2.10 m high and 12.4 m long. At the bottom

babymother [125]

Answer:

630.75 j

Explanation:

from the question we have the following

total mass (m) = 54.5 kg

initial speed (Vi) = 1.4 m/s

final speed (Vf) = 6.6 m/s

frictional force (FF) = 41 N

height of slope (h) = 2.1 m

length of slope (d) = 12.4 m

acceleration due to gravity (g) = 9.8 m/s^2

work done (wd) = ?

we can calculate the work done by the boy in pushing the chair using the law of law of conservation of energy

wd + mgh = (0.5 mVf^2) - (0.5 mVi^2) + (FF x d)

wd = (0.5 mVf^2) - (0.5 mVi^2) + (FF x d) - (mgh)

where wd = work done

m = mass

h = height

g = acceleration due to gravity

FF = frictional force

d = distance

Vf and Vi = final and initial velocity

wd = (0.5 x 54.5 x 6.9^2) - (0.5 x 54.5 x 1.4^2) + (41 x 12.4) - (54.5 X 9.8 X 2.1)

wd = 630.75 j

3 0

3 years ago

A meteoroid is traveling east through the atmosphere at 18. 3 km/s while descending at a rate of 11.5 km/s. What is its speed, i

Annette [7]

Answer:

The speed of meteoroid is 21.61 km/s in south-east.

Explanation:

Given that,

A meteoroid is traveling through the atmosphere at 18.3 km/s. while descending at a rate of 11.5 km/s it means 11.5 km/s in south.

We need to draw a diagram

Using Pythagorean theorem

$AC^2=AB^2+BC^2$

$AC^2=(18.3)^3+(11.5)^2$

$AC=\sqrt{(18.3)^2+(11.5)^2}$

$AC=21.61\ km/s$

Hence, The speed of meteoroid is 21.61 km/s in south-east.

6 0

3 years ago

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

spayn [35]

Answer:

$v_0 = 3.53~{\rm m/s}$

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

$x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}$

For the y-direction gives

$v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t$

Combining both equation yields the y_component of the final velocity

$v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}$

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

$\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}$

6 0

3 years ago

A fisherman is fishing from a bridge and is using a "42.0-N test line." In other words, the line will sustain a maximum force of

lara31 [8.8K]

Answer:

(a) 42 N

(b)36.7 N

Explanation:

Nomenclature

F= force test line (N)

W : fish weight (N)

Problem development

(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed

We apply Newton's first law of equlibrio because the system moves at constant speed:

∑Fy =0

F-W= 0

42N -W =0

W = 42N

(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²

We apply Newton's second law because the system moves at constant acceleration:

m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity

∑Fy =m*a

m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity

F-W= ( W/9.8 )*a

42-W= ( W/9.8 )*1.41

42= W+0.1439W

42=1.1439W

W= 42/1.1439

W= 36.7 N

8 0

3 years ago