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devlian
1 month ago
9

A mouse runs along a baseboard in your house. The mouse's position as a function of time is given by x(t)=pt2+qt, with p = 0.36

m/s2 and q = -1.20 m/s .
Determine the mouse's average speed between t=1.0s and t=4.0s
Physics
1 answer:
ValentinkaMS [3.4K]1 month ago
5 0

Response:

0.60 m/s

Details:

The average speed between times t = a and t = b can be expressed as:

v_avg = (x(b) − x(a)) / (b − a)

Given the function x(t) = 0.36t² − 1.20t, and considering the interval from 1.0 to 4.0:

v_avg = (x(4.0) − x(1.0)) / (4.0 − 1.0)

v_avg = [(0.36(4.0)² − 1.20(4.0)) − (0.36(1.0)² − 1.20(1.0))] / 3.0

v_avg = [(5.76 − 4.8) − (0.36 − 1.20)] / 3.0

v_avg = [0.96 − (-0.84)] / 3.0

v_avg = 0.60

The average speed calculated is 0.60 m/s.

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inna [3103]

The peak wavelength for Betelgeuse is 828 nm

Explanation:

Wien's law describes how the surface temperature relates to a star’s peak wavelength:

\lambda=\frac{b}{T}

where

\lambda represents the peak wavelength

T is the surface temperature

b=2.898\cdot 10^{-3} m\cdot Kis Wien's constant

For Betelgeuse, the surface temperature is roughly

T = 3500 K

Consequently, its peak wavelength can be determined as:

\lambda=\frac{2.898\cdot 10^{-3}}{3500}=8.28\cdot 10^{-7} m = 828 nm

Learn more about wavelength:

8 0
1 month ago
A small mass m is tied to a string of length L and is whirled in vertical circular motion. The speed of the mass v is such that
serg [3582]

Response:

(mv^2/R)/(mg)=1/2

v^2=R/2g

7 0
1 month ago
On its own, a certain tow-truck has a maximum acceleration of 3.0 m/s2. what would be the maximum acceleration when this truck w
inna [3103]

Let us denote a_1=3 m/s^2 as the highest acceleration of the truck by itself. The force generated by the engine to propel the truck in this scenario is
F= ma_1
where m is the mass of the truck alone.

Upon attaching a bus that has double the mass of the truck, the total mass of the entire system (truck+bus) becomes (m+2m)=3m. In this situation, the force generated by the engine is
F=3m a_2
where a2 represents the new acceleration. 
Since the engine remains unchanged, the force generated stays the same, allowing us to set the force equations equal to each other:
m a_1 = 3 m a_2
and solving for a2 gives us:
a_2 = \frac{m a_1}{3 m}= \frac{a_1}{3}= \frac{3 m/s^2}{3}=1 m/s^2

5 0
1 month ago
Read 2 more answers
A 7.5 kg cannon ball leaves a canon with a speed of 185 m/s. Find the average net force applied to the ball if the cannon muzzle
Keith_Richards [3271]

To determine the average net force, we can calculate acceleration using:

x = 0.5*a*t^2

v = a*t

where x=3.6m and v=185 m/s.

Thus,

t=v/a and therefore x = 0.5*a*(v/a)^2 = 0.5 * (v^2)/a

which gives us a= (0.5*v^2)/x

Since we have the known values of v and x, we can compute a by substituting these numbers.

The average net force is then given as:

F = m*a,

with m=7.5kg.


5 0
1 month ago
Two projectiles are launched with the same initial speed from the same location, one at a 30° angle and the other at a 60° angle
Ostrovityanka [3204]

Answer:

Explanation:

Let us denote the launch angle as \theta _1=30^{\circ}.

\theta _2=60^{\circ}, \theta _1, and \theta _2 are complementary angles which means their ranges are identical.

H_{max}=\frac{u^2(\sin \theta )^2}{2g}

H_{30}=\frac{u^2(\sin 30)^2}{2g}

H_{30}=\frac{u^2}{8g}

Time of flight is indicated as =\frac{2u\sin \theta }{g}.

t_{30}=\frac{u}{g}

For \theta =60^{\circ}

H_{60}=\frac{u^2(\sin 60)^2}{2g}

H_{60}=\frac{3u^2}{8g}

t_{60}=\frac{2u\sin 60}{g}=\frac{\sqrt{3}u}{g}

H_{60}=3\times H_{30}

t_{60}=\sqrt{3}\times t_{30}

7 0
1 month ago
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