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Airida [17]
2 years ago
15

ME PUEDEN DECIR EJERCICIOS DE RESISTENCIA AÉROBICOS Y ANAERÓBICOS POR FAAA :C ES URGENTE *Me lo explican bien pliis para saber k

omo hacerlo :)
Physics
1 answer:
Anvisha [2.4K]2 years ago
6 0

Explanation:

sksksksmsmamamamamamamamama

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What is the substance through which sound travels?
Ann [662]

Answer:

Sound waves need to travel through a medium such as solids, liquids and gases. The sound waves move through each of these mediums by vibrating the molecules in the matter. The molecules in solids are packed very tightly. Liquids are not packed as tightly.

Explanation:

Hope this helped, Have a Great Day!!

4 0
2 years ago
The specific heat of acetic acid is 2.07 J/gºC. If 1150 J of heat is
Doss [256]

Answer:

12.3 °C

Explanation:

Q=m*Cp*ΔT-->

1150=45*2.07*ΔT-->

ΔT=1150/93.15-->

ΔΤ=12.3 °C

4 0
3 years ago
A solid ball and a hollow ball, each with a mass of 1.00 kg and radius of 0.100 m start from rest and roll down a ramp of length
nordsb [41]

Answer:

The solid ball and hollow ball both will reach the bottom with the same speed.

Explanation:

The speed of the solid and hollow balls is independent of  the mass and the radius. A solid and hollow ball experience same speed on a given incline.

The speed can be calculated as

v = √(10/7)gh

where g is gravitational acceleration and h is the height

sinθ = h/L

h = L*sinθ

h = 3*sin(35)

h = 1.72 m

v = √(10/7)*9.8*1.72

v = 4.91 m/s

Both balls will reach the bottom at the speed of 4.91 m/s.

8 0
3 years ago
A team of engineering students is designing a catapult to launch a small ball at A so that it lands in the box. If it is known t
Murljashka [212]

Answer:

Explanation:

If the initial velocity is U

Then the horizontal component of the velocity is

Ux= Ucosθ

Then the range for a projectile is give as

R=Ux.t

Where t is the time of flight

The time of flight is given as

t=2USinθ/g

Therefore,

R=Ux.t

R=UCosθ.2USinθ/g

R=U^2×2SinθCosθ/g

Then, from trigonometric ratio

2SinθCosθ= Sin2θ

R=U^2Sin2θ/g

Given that θ=32° and g=9.81m/s^2

Then

R=U^2Sin2×32/9.81

R=U^2Sin64/9.81

R=0.0916U^2

Then, range is given by R=0.0916U^2

A=0.0916U^2.

T

The box is at a distance A from the point of projection. Then the range R=A

R=0.0916U^2

A=0.0916U^2

Then,

U^2=A/0.0916

U^2=10.915A

Then the initial velocity should be

U=√10.915A

U=3.3√A

8 0
3 years ago
A client with hypertension who weighs 72.4 kg is receiving an infusion of nitroprusside (Nipride) 50 mg in D5W 250 ml at 75 ml/h
Mkey [24]

To solve this problem it is necessary to simply apply the concepts related to cross-multiply and proportion between units.

Let's start first by relating the amount of dose needed to be supplied per hour, in other words,

The infusion of 250ml should be supplied at a rate of 75ml / hour, so what amount x of mg hour should be supplied with 50Mg.

\frac{x}{75ml/hour} \rightarrow \frac{50mg}{250ml}

x \rightarrow \frac{50mg*75ml/hour}{250ml}

x \rightarrow \frac{3750mg}{250hour}

x \rightarrow 15\frac{mg}{hour}

Converting to mcg units we know that 1mg is equal to 1000mcg and that 1 hour contains 60 min, therefore

x \rightarrow 15\frac{mg}{hour}

x \rightarrow 15\frac{mg}{hour}(\frac{1000mcg}{1mg})(\frac{1hour}{60min})

x \rightarrow 250mcg/min

The dose should be distributed per kilogram of the patient so if the patient weighs 72.4kg,

Dose = \frac{250mcg/min}{72.4kg}

Dose = 3.5 \frac{mcg/min}{kg}

Therefore the client will receive 3.5mcg/kg/min.

8 0
3 years ago
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