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2 years ago

12

Una banda transportadora se utiliza para cargar canastos en un avión de carga, si µs = 0.5. ¿Cuál es el ángulo máximo de elevaci

ón para evitar el deslizamiento? Respuesta: 26.6

1 answer:

Sonja [21]2 years ago

7 0

No se creo es 45.7 de nada

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The attractive electrostatic force between the point charges +8.44 ✕ 10-6 and q has a magnitude of 0.961 n when the separation b

d1i1m1o1n [39]

The electrostatic force between two charges Q1 and q is given by
$F=k_e \frac{Q_1 q}{r^2}$
where
ke is the Coulomb's constant
Q1 is the first charge
q is the second charge
r is the distance between the two charges

Re-arranging the formula, we have
$q= \frac{F r^2}{k_e Q_1}$
and since we know the value of the force F, of the charge Q1 and the distance r between the two charges, we can calculate the value of q:
$q= \frac{(0.961 N)(0.67 m)^2}{(8.99 \cdot 10^9 N m^2 C^{-2})(+8.44\cdot 10^{-6}C)}=5.69 \cdot 10^{-6} C$

And since the force is attractive, the two charges must have opposite sign, so the charge q must have negative sign.

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3 years ago

Relate RNA to translation and transcription

vlabodo [156]

Answer:

RNA plays a role in both Translation and transcription.

Explanation:

RNA translation uses mRNA (or messenger RNA) in order to bring the coding to the ribosome.

RNA transcription is where the coding receives a new strand of coding (replacing thymine with uracil) , which is then turned into protein

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2 years ago

A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f

konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

$A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m$

(c) the maximum speed attained by the object during its motion

$E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s$

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2 years ago

The total distance from a house to a school to is 9.5 km. A student travels all the way from his house to the school and back to

marysya [2.9K]

Answer:

0

Explanation:

Displacement is a vector from initial to final point. Because initial and final point are the same, so displacement is 0.

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2 years ago

Từ độ cao 100 m người ta thả một vật thẳng đứng xuống với v = 10 m/s, g = 10 m/s2 . a. Sau bao lâu vật chạm đất. b. Tính vận tốc

Bas_tet [7]

Answer:

10 points dapat yawa aman daw gud

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2 years ago

Read 2 more answers