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3 years ago

What muscle will Lunges work ? A) Bicep B)Tricep C) quadriceps D) pectoralis major

Physics

1 answer:

ivann1987 [24]3 years ago

8 0

Answer:

C. quadriceps

Explanation:

I played Ring Fit adventure and I have some experience with the exercises. It tells you which muscle you are working when you do a move, and I remember the lunge worked the quadriceps!

Plz mark brainliest if right and thank you!

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A cylindrical Nickel rod (9 mm diameter, 50 m long) is pulled in tension with a load of 6,283 N. What would the elongation of th

Norma-Jean [14]

Answer:

0.29 m

Explanation:

9 mm = 0.009 m in diameter

Cross-sectional area $A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36\times 10^{-5} m^2$

Let the tensile modulus of Nickel $E = 170 \times 10^9Pa$ .

The elongation of the rod can be calculated using the following formula:

$\Delta L = \frac{F L}{A E} = \frac{6283*50}{6.36\times 10^{-5} * 170 \times 10^9} = \frac{314150}{1081200} = 0.29 m$

6 0

4 years ago

What causes weather to firm into droplets.

Tomtit [17]

Answer:

Density

Explanation:

Don't @ me lol

6 0

3 years ago

A lightning flash releases about 1010J of electrical energy. Part A If all this energy is added to 50 kg of water (the amount of

zavuch27 [327]

Answer:

water is in the vapor state,

Explanation:

We must use calorimetry equations to find the final water temperatures. We assume that all energy is transformed into heat

E = Q₁ + $Q_{L}$

Where Q1 is the heat required to bring water from the current temperature to the boiling point

Q₁ = m $c_{e}$ ( $T_{f}$ -T₀)

Q₁ = 50 4180 (100 - 37)

Q₁ = 1.317 10⁷ J

Let's calculate the energy so that all the water changes state

$Q_{L}$ = m L

$Q_{L}$ = 50 2,256 106

$Q_{L}$ = 1,128 10⁸ J

Let's look for the energy needed to convert all the water into steam is

Qt = Q₁ + $Q_{L}$

Qt = 1.317 107 + 11.28 107

Qt = 12,597 10⁷ J

Let's calculate how much energy is left to heat the water vapor

ΔE = E - Qt

ΔE = 10¹⁰ - 12,597 10⁷

ΔE = 1000 107 - 12,597 107

ΔE = 987.4 10⁷ J

With this energy we heat the steam, clear the final temperature

Q = ΔE = m $c_{e}$ ( $T_{f}$ -To)

( $T_{f}$ -T₀) = ΔE / m $c_{e}$

$T_{f}$ = T₀ + ΔE / m $c_{e}$

$T_{f}$ = 100 + 987.4 10⁷ / (50 1970)

$T_{f}$ = 100 + 1,002 10⁵

$T_{f}$ = 1,003 10⁵ ° C

This result indicates that the water is in the vapor state, in realizing at this temperature the water will be dissociated into its hydrogen and oxygen components

4 0

3 years ago

The cost of energy delivered to residences by electrical transmission varies from $0.070/kWh to $0.258/kWh throughout the United

tankabanditka [31]

Answer:

(a) $ 1.48

(b) $ 0.005

Explanation:

(a)

First, we calculate the energy consumed, by using following formula:

E = Pt

where,

E = Energy Consumed = ?

P = Power Delivered = (40 W)(1 KW/1000 W) = 0.04 KW

t = Time Duration = (2 weeks)(7 days/week)(24 h/day) = 336 h

Therefore,

E = (0.04 KW)(336 h) = 13.44 KWh

Now, we calculate cost, by following formula:

Cost = (E)(Unit Price)

Cost = (13.44 KWh)($ 0.11/KWh)

Cost = $ 1.48

(b)

First, we calculate the energy consumed, by using following formula:

E = Pt

where,

E = Energy Consumed = ?

P = Power Delivered = (970 W)(1 KW/1000 W) = 0.97 KW

t = Time Duration = (3 min)(1 h/60 min) = 0.05 h

Therefore,

E = (0.97 KW)(0.05 h) = 0.0485 KWh

Now, we calculate cost, by following formula:

Cost = (E)(Unit Price)

Cost = (0.0485 KWh)($ 0.11/KWh)

Cost = $ 0.005

(c)

First, we calculate the energy consumed, by using following formula:

E = Pt

where,

E = Energy Consumed = ?

P = Power Delivered = (5.203 x 10³ W)(1 KW/1000 W) = 5.203 KW

t = Time Duration = (40 min)(1 h/60 min) = 0.67 h

Therefore,

E = (5.203 KW)(0.67 h) = 3.47 KWh

Now, we calculate cost, by following formula:

Cost = (E)(Unit Price)

Cost = (3.47 KWh)($ 0.11/KWh)

Cost = $ 0.38

8 0

3 years ago

If a sound is 30 dB and its absolute pressure was 66 x 10-9 Pa, what must have been the reference pressure?

Ksenya-84 [330]

Given:

I = 30dB

P = 66 × $10^{-9}$ Pa

Solution:

Formula used:

I = $20\log_{10}(\frac{P}{P_{o}})$ (1)

where,

I = intensity of sound

P = absolute pressure

$P_{o}$ = reference pressure

Using Eqn (1), we get:

$30 = 20\log _{10}\frac{66\times 10^{-9}}{P^{o}}$

$P_{o}$ = $\frac{66\times 10^{-9}}{10^{1.5}}$

$P_{o}$ = 2.08 × $10^{-9}$ Pa

4 0

4 years ago