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MA_775_DIABLO [31]
3 years ago
13

A sled of mass m is given a kick on a frozen pond. The kick imparts to it an initial speed of 2.20 m/s. The coefficient of kinet

ic friction between sled and ice is 0.115. Use energy considerations to find the distance the sled moves before it stops. m
Physics
1 answer:
Makovka662 [10]3 years ago
3 0

Answer:

2.15 m

Explanation:

u = 2.2 m/s, v = 0, uk = 0.115

a = uk x g = 0.115 x 9.8 = 1.127 m/s^2

Let s be the distance traveled before stopping.

v^2 = u^2 - 2 a s

0 = 2.2^2 - 2 x 1.127 x s

s = 2.15 m

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Answer:

no

Explanation:

they cannot because they contain the same amount of liquid

8 0
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When an object moves, stops moving, changes speed, or changes direction, how do scientists describe that condition?
lorasvet [3.4K]
Drop "moves" from the list for a moment.

You can also drop "stops moving", because that's included in "changes speed"
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When an object changes speed or changes direction, that's called "acceleration".

I dropped the first one from the list, because an object can be moving,
and as long as it's speed is constant and it's moving in a straight line,
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I think you meant to say "starts moving". That's a change of speed (from zero
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8 0
3 years ago
If u drop a bar of soap on the floor then is the floor clean or is the bar of soap dirty? Lol I’m so bored i dont even know what
emmainna [20.7K]

Answer:

CLEAN

Explanation:

6 0
3 years ago
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An electric bulb is marked 40volts ,230w another bulb is marked 40w,110v
Andrej [43]

Answer:

a. The ratio of their resistance is 2783:64

b. The ratio of their energy is 4:23

c. The charge on the first bulb is 5.75 C

The charge on the second bulb is 0.\overline {36} C

Explanation:

The voltage on one of the electric bulbs, V₁ = 40  volts

The power rating of the bulb, P₁ = 230 w

The voltage on the other electric bulbs, V₂ = 110 volts

The power rating of the bulb, P₂ = 40 w

a. The power is given by the formula, P = I·V = V²/R

Therefore, R = V²/P

For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96

The resistance of the second bulb, R₂ = 110²/40

The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64

∴ The ratio of their resistance, R₂:R₁ = 2783:64

b. The energy of a bulb, E = t × P

Where;

t = The time in which the bulb is powered on

∴ The energy of the first bulb, E₁ = 230 w × t

The energy of the second bulb, E₂ = 40 w × t

The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23

∴ The ratio of their energy, E₂:E₁ = 4:23

c. The charge on a bulb, 'Q', is given by the formula, Q = I × t

Where;

I = The current flowing through the bulb

From P = I·V, we get;

I = P/V

For the first bulb, the current, I = 230 w/40 V = 5.75 amperes

The charge on the first bulb per second (t = 1) is therefore;

Q₁ = 5.75 A × 1 s = 5.75 C

The charge on the first bulb, Q₁ = 5.75 C

Similarly, the charge on the second bulb, Q₂ = (40 W/110 V) × 1 s = 0.\overline {36} C

The charge on the second bulb, Q₂ = 0.\overline {36} C.

d. The question has left out parts

4 0
3 years ago
A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness
Mekhanik [1.2K]

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

\sigma_h = \frac{Pd}{2t}

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

\sigma_l = \frac{Pd}{4t}

The letters have the same meaning as before.

Then he hoop stress would be,

\sigma_h = \frac{Pd}{2t}

\sigma_h = \frac{75 \times 8}{2\times 0.25}

\sigma_h = 1200psi

And the longitudinal stress would be

\sigma_l = \frac{Pd}{4t}

\sigma_l = \frac{75\times 8}{4\times 0.25}

\sigma_l = 600Psi

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

\tau_{max} = \frac{\sigma_h}{2}

\tau_{max} = \frac{1200}{2}

\tau_{max} = 600Psi

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

6 0
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