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Bingel [31]
3 years ago
13

A bumblebee is flying to the right when a breeze causes the bee to slow down with a constant leftward acceleration of magnitude

0.50\,\dfrac{\text m}{\text s^2}0.50 s 2 m ​ 0, point, 50, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction. After 2.0\,\text{s}2.0s2, point, 0, start text, s, end text, the bee is moving to the right with a speed of 2.75\,\dfrac{\text m}{\text s}2.75 s m ​ 2, point, 75, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. What was the velocity of the bumblebee right before the breeze?
Physics
1 answer:
lesya692 [45]3 years ago
7 0

Answer:

3.75 m/s

Explanation:

Given:

v = 2.75 m/s

a = -0.50 m/s²

t = 2.0 s

Find: v₀

v = at + v₀

2.75 m/s = (-0.50 m/s²) (2.0 s) + v₀

v₀ = 3.75 m/s

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The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC
enyata [817]

Answer:

The y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

Explanation:

<u>Given:</u>

  • Electric field in the region, \vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.
  • Charge placed into the region, q = 10.4\ nC = 10.4\times 10^{-9}\ C.

where, \hat i,\ \hat j are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.

Thus, the y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

3 0
3 years ago
A person in bare feet is standing under a tree during a thunderstorm, seeking shelter from the rain. A lightning strike hits the
Anni [7]

Answer:

I=38.181\ A is the current through the body of the man.

E=34.5\ J energy dissipated.

Explanation:

Given:

  • time for which the current lasted, t=43\times 10^{-6}\ s
  • potential difference between the feet, V=21000\ V
  • resistance between the feet, R=550\ \Omega

<u>Now, from the Ohm's law we have:</u>

I=\frac{V}{R}

I=\frac{21000}{550}

I=38.181\ A is the current through the body of the man.

<u>Energy dissipated in the body:</u>

E=I^2.R.t

E=38.181^2\times 550\times 43\times 10^{-6}

E=34.5\ J

5 0
3 years ago
The magnetic flux through each turn of a 110-turn coil is given by ΦB = 9.75 ✕ 10−3 sin(ωt), where ω is the angular speed of the
Xelga [282]

Answer:

Explanation:

Given that a coil has a turns of

N = 110 turns

And the flux is given as function of t

ΦB = 9.75 ✕ 10^-3 sin(ωt),

Given that, at an instant the angular velocity is 8.70 ✕ 10² rev/min

ω = 8.70 ✕ 10² rev/min

Converting this to rad/sec

1 rev = 2πrad

Then,

ω = 8.7 × 10² × 2π / 60

ω = 91.11 rad/s

Now, we want to find the induced EMF as a function of time

EMF is given as

ε = —NdΦB/dt

ΦB = 9.75 ✕ 10^-3 sin(ωt),

dΦB/dt = 9.75 × 10^-3•ω Cos(ωt)

So,

ε = —NdΦB/dt

ε = —110 × 9.75 × 10^-3•ω Cos(ωt)

Since ω = 91.11 rad/s

ε = —110 × 9.75 × 10^-3 ×91.11 Cos(91.11t)

ε = —97.71 Cos(91.11t)

The EMF as a function of time is

ε = —97.71 Cos(91.11t)

Extra

The maximum EMF will be when Cos(91.11t) = -1

Then, maximum emf = 97.71V

8 0
3 years ago
Can anyone please tell me where these labels would go?
Nadya [2.5K]
The least potential energy would go at the very bottom of the track. the greatest kinetic energy would be on the upper half of the track and the least kinetic energy would be on the lower half of the track. please review this on google if you are not sure.
6 0
3 years ago
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Bob is cleaning his dishwasher with white vinegar by putting a bowl of this ____________ mixture in the top of an empty dishwash
Maurinko [17]

Answer: Baking Soda

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7 0
3 years ago
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