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3 years ago

13

A rectangle had sides of 8.80 ft and 24.1 ft. Calculate the area of the rectangle in cm^2 (you should get 197000 cm^2) show your

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1 answer:

Ratling [72]3 years ago

4 0

I ft = 30.48 cm

length l = 8.80 ft = 8.80* (30.48 cm) = 268.224 cm

width w = 24.1 ft = 24.1 * (30.48 cm) = 734.568 cm

Area A = l*w = (268.224 cm)(734.568 cm)

A = 197028 cm^2

A = 197000 cm^2

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What is the mass of 2000 ml of an intravenous glucose solution with a density of 1.15 g/ml?

Scorpion4ik [409]

According to the following formula, the answer is 2,300 g or 2.3 kg:

Volume (m)/Mass (m) Equals Density (p) (V)

Here, the density is 1.15 g/mL, allowing the formula described above to result in a mass of 2.00 L:

p=m/V

1.15 g/mL is equal to x g/2.00 L or x g/2,000 mL.

2,000 mL of x g = 1.15 g of g/mL

2.3 kg or 2,300 g for x g.

<h3>How many grams of glucose are in a 1000ml bag of glucose 5?</h3>

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5 0

1 year ago

A 1.2-kg mass suspended from a spring of spring constant 22 N.m-1 executes simple harmonic motion of amplitude 5 cm. What is the

iren2701 [21]

Answer:

a) T = 1,467 s , b) A = 0.495 m , c) v = 4.97 10⁻² m / s

Explanation:

The simple harmonic movement is described by the expression

x = A cos (wt + Ф)

Where the angular velocity is

w = √ k / m

a) Ask the period

Angular velocity, frequency and period are related

w = 2π f = 2π / T

T = 2π / w

T = 2pi √ m / k

T = 2π √ (1.2 / 22)

T = 1,467 s

f = 1 / T

f = 0.68 Hz

b) ask the amplitude

The mechanical energy of a harmonic oscillator

E = ½ k A²

A = √2 E / k

A = √ (2 2.7 / 22)

A = 0.495 m

c) the mass changes to 8.0 kg

As released from rest Ф = 0, the equation remains

x = A cos wt

w = √ (22/8)

w = 1,658

x = 3.0 cos (1,658 t)

Speed is

v = dx / dt

v = -A w sin wt

The speed is maximum when without wt = ±1

v = Aw

v = 0.03 1,658

v = 4.97 10⁻² m / s

6 0

2 years ago

The density of atmosphere on a certain planet is found to decrease as altitude increases what is the type of relationship betwee

8_murik_8 [283]

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6 0

3 years ago

(a) Find the frequency of revolution of an electron with an energy of 114 eV in a uniform magnetic field of magnitude 46.7 µT. (

stira [4]

Answer:

(a) 1.3 x 10^6 Hz

(b) 76.73 cm

Explanation:

(a)

the formula for the frequency is given by

f = B q / 2 π m

where, B be the strength of magnetic field, q be the charge on one electron, m is the mass of one electron.

B = 46.7 micro tesla = 46.7 x 10^-6 T

q = 1.6 x 10^-19 C

m = 9.1 x 10^-31 kg

f = (46.7 x 10^-6 x 1.6 x 10^-19) / (2 x 3.14 x 9.1 x 10^-31) = 1.3 x 10^6 Hz

(b) K = 114 eV = 114 x 1.6 x 10^-19 J = 182.4 x 10^-19 J

K = 1/2 mv^2

182.4 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2

v = 6.3 x 10^6 m/s

r = m v / B q

Where, r be the radius of circular path

r = (9.1 x 10^-31 x 6.3 x 10^6) / (46.7 x 10^-6 x 1.6 x 10^-19)

r = 0.7673 m = 76.73 cm

5 0

3 years ago

A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42◦ from t

Semmy [17]

Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

P.E = mgh joule

Considering h = 0 for the vertical position

The h at ∅ = 42° is h = L (1 - cos∅)

P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

P.E = 25 x 9.8 x 2.2 (1 - cos42°)

= 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

∴ K.E = P.E

1/2 mv² = 138.44

1/2 x 25 x v² 138.44

v² = 11.0752

v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

Tension on string, T = Force acting on the swing, F

$W=L\int\limits^0_\phi{F} \, d \phi$

$=L\int\limits^0_\phi{mg.sin \phi} \, d \phi$

= $-Lmg[cos\phi]_{42}^{0}$

= - 2.2 x 25 x 9.8 [cos0 - cos 42°]

= - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

3 0

3 years ago