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Scilla [17]
3 years ago
12

3 + 2 ∙ 4 = 3 + (2 ∙ 4)

Physics
2 answers:
Rzqust [24]3 years ago
6 0

Answer:

11 = 11

Explanation:

AnnZ [28]3 years ago
6 0

Answer:

True

Explanation:

3+2x4=3+(2x4), they both equal 11

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Calculate the percentage increase in speed of the cyclist when the power output changes from 200 W to 300 W.
Rom4ik [11]

Answer:

<em>50%</em>

Explanation:

Given

Initial power = 200W

Final power = 300W

Increment = 300 - 200 = 100W

percentage increase = increment/initial power * 100

percentage increase = 100/200 * 100%

percentage increase = 0.5 * 100

percentage increase = 50%

<em>Hence the percentage increase in speed is 50%</em>

6 0
2 years ago
A typical land snail's speed is 12.2 meters per hour. How many miles will the snail travel in one day(24hrs)?
algol13
The snail will go <span>0.18193752 miles </span>
5 0
2 years ago
An object, initially at rest, is subject to an acceleration of 34 m/s2. How long will it take for that
SIZIF [17.4K]

Answer:

If it is moving 34 m/s it will take 100 seconds, or 1:40 to reach 3400 meters.

Explanation:

I found this answer by dividing 3400 by 34 and converting seconds to minutes

6 0
3 years ago
Read 2 more answers
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
2 years ago
Match the following properties to the type of wave.
Reil [10]

Answer:

hi there

Explanation:

1 - III

2- 1

3-1

HOPE IT HELP YOU

PLz mark me as a BRAINLIST

6 0
2 years ago
Read 2 more answers
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