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dalvyx [7]
3 years ago
9

3. During a race, a sprinter increases from 5.0 m/s to 7.5 m/s over a period of 1.25s. What is the sprinter’s average accelerati

on during this period?
Also please try and explain
Please and Thank you
Physics
1 answer:
Rasek [7]3 years ago
5 0
(7.5)-(5)
----------- = 2 m/s^2
(1.25-0)

Average acceleration is calculated using the equation V(f)-V(i) / t(f)-t(i). (Final velocity minus initial velocity divided by final time minus initial time)
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What is 0.94kg divided by 2.4n
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A stuffed toy with a mass of 0.900 kilograms sits on the edge of a bed at a height of 0.830 if the toy falls off the bed what wi
Olenka [21]
Mechanical energy (ME) is the sum of potential energy (PE) and kinetic energy (KE). When the toy falls, energy is converted from PE to KE, but by conservation of energy, ME (and therefore PE+KE) will remain the same.

Therefore, ME at 0.500 m is the same as ME at 0.830 m (the starting point). It's easier to calculate ME at the starting point because its just PE we need to worry about (but if we wanted to we could calculate the instantaneous PE and KE at 0.500 m too and add them to get the same answer).

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8 0
3 years ago
Particle q₁ has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10⁻³ N, what is the s
Ne4ueva [31]
The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
F=qvB \sin \theta
where \theta is the angle between the directions of v and B.

1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is q=2.7 \mu C=2.7 \cdot 10^{-6}C. In this case, v and B are perpendicular, so \theta=90^{\circ}, therefore we have:
B= \frac{F}{qv \sin \theta} = \frac{5.75 \cdot 10^{-3}N}{(2.7 \cdot 10^{-6}C)(773m/s)\sin 90^{\circ}}=2.8 T

2) In this second case, the angle between v and B is \theta=55^{\circ}. The charge is now q=42.0 \mu C=42.0 \cdot 10^{-6}C, and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
F=qvB \sin \theta=(42\cdot 10^{-6}C)(1.21 \cdot 10^3 m/s)(2.8 T)(\sin 55^{\circ})=0.12 N
5 0
3 years ago
Read 2 more answers
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