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3 years ago

A motorcycle drives up a steeply inclined ramp. The work done on the motorcycle by the earth’s gravitational force is ____.

Physics

2 answers:

sergejj [24]3 years ago

4 0

Answer: The work is negative

Explanation: The work is defined as the movement done by a force.

In this case, becuse the motocycle is driving upward in a inclinated ramp, the force of gravity will pull downwards, in the oposite direction of wich the motocycle is moving, this means that the work is negative, this means that the force is oposing to the motion of the motocycle and the motocycle needs to cancel this negative work in order to move.

PilotLPTM [1.2K]3 years ago

3 0

Negative - when the motorcycle displacement has a component in the opposite direction of the given force, that force is doing negative work on the motorcycle.

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A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west

ivolga24 [154]

Answer:

Magnitude of avg velocity, $|v_{avg}| = 18.9 km/h$

$\theta' = 56.85^{\circ}$

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = $\frac{27}{60} h$

Angle, $\theta = 50^{\circ}$

Time taken in $50^{\circ}$ east of due North direction, t' = 29 min = $\frac{29}{60} h$

Time taken in west direction, t'' = 37 min = $\frac{27}{60} h$

Solution:

Now, the displacement, 's' in east direction is given by:

$\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km$

Displacement in $50^{\circ}$ east of due North for 29.0 min is given by:

$\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}$

$\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}$

$\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km$

Now, displacement in the west direction for 37 min:

$\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km$

Now, the overall displacement,

$\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}$

$\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}$

$\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km$

(a) Now, average velocity, $v_{avg}$ is given:

$v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}$

$v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}$

$v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h$

Magnitude of avg velocity is given by:

$|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h$

(b) angle can be calculated as:

$tan\theta' = \frac{15.83}{10.34}$

$\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}$

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3 years ago

How much positive electric charge is in 10 moles of carbon? NA=6.022×1023mol−1, e=1.60×10−19C.

levacccp [35]

Answer:

4.75 x 10^-25 kg

Explanation:

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2 years ago

In the first-order spectrum, maxima for two different wavelengths are separated on the screen by 3.40 mm . what is the differenc

Strike441 [17]

The solution would be like this for this specific problem:

Given:

diffraction grating slits = 900 slits per centimeter

interference pattern that is observed on a screen from the grating = 2.38m

maxima for two different wavelengths = 3.40mm

slit separation .. d = 1/900cm = 1.11^-3cm = 1.111^-5 m 

Whenas n = 1, maxima (grating equation) sinθ = λ/d
Grant distance of each maxima from centre = y ..
As sinθ ≈ y/D y/D = λ/d λ = yd / D 

∆λ = (λ2 - λ1) = y2.d/D - y1.d/D
∆λ = (d/D) [y2 -y1]

∆λ = 1.111^-5m x [3.40^-3m] / 2.38m .. .. ►∆λ = 1.587^-8 m