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MrRissso [65]
2 years ago
14

What time does the clock go back for daylight savings

Physics
1 answer:
Novosadov [1.4K]2 years ago
6 0

Answer:

it goes back an hour from the time it was

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Why does it dangerous to jump from a moving vehicle?
Kaylis [27]
Why is it dangerous*btw.....And because you can easily crash into another vehicle also because it's just plain stupid to jump anyway ...
4 0
3 years ago
a deer with a mass of 176 kg is running head-on towards you with a velocity of 19 m/s. you are going north. find the magnitude a
Igoryamba

Momentum = Mass × Velocity

According to this formula,

Momentum of deer = 176 × 19 = 3344 kg•m/s.

Since you are heading north and the deer is running towards you, the direction of the deer' s momentum is north as well.

6 0
3 years ago
A mass of 2 kg traveling at 3 m/s undergoes a one-dimensional elastic collision with a group of four 1kg masses that are at rest
alukav5142 [94]
2kg + 3 kg = 5 kg as result of the conclusion
6 0
3 years ago
Joshua was driving to a friend’s house to study. During his trip, he started on pavement. At one point, he hit an ice patch on t
Tems11 [23]

Answer:

b. Friction decreased when he went from pavement to ice and then increased two more times.

Explanation:

Frictional force depends on the normal force of the surface and a friction coefficient.

F_{f} = -\mu N

Since we're talking about the same car, the value of N will remain constant whereas μ will represent the change in the frictional coefficient of the surface. Now we consider the different surfaces, cars will slide in an icy road which means that the frictional coefficient is smaller than the pavement.

After Joshua returns to the pavement road, the resulting frictional force increases and will do so one more time when he reaches the gravel road. Gravel roads have greater frictional coefficients than pavement roads which means the frictional force will increase a second time.

7 0
3 years ago
Read 2 more answers
A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with
Julli [10]

Explanation:

Given that,

Capacitor C=1.66\ \mu F

Resistor R=80.0\ \Omega

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

f_{c}=\dfrac{1}{2\pi R C}

Where, R = resistor

C = capacitor

Put the value into the formula

f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}

f_{c}=1198.45\ Hz

(B). We need to calculate the V_{R} when f = \dfrac{1}{2f_{c}}

Using formula of  V_{R}

V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})

Put the value into the formula

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=2.280\ Volt

(C). We need to calculate the V_{R} when f = f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=3.606\ Volt

(D). We need to calculate the V_{R} when f = 2f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=4.561\ Volt

Hence, This is the required solution.

8 0
3 years ago
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