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3 years ago

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¿Qué velocidad inicial debe impartirse a una masa de 5kg para elevarla a una altura de 10m? ¿Cuál es la energía total en cualqui

era de los puntos de su trayectoria?

1 answer:

Snowcat [4.5K]3 years ago

6 0

Cada 100 cm es de 1 m por lo que tomar los kilos a su vez a cm, a continuación, a su vez a metros

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The towline exerts a force of p = 4 kn at the end of the 20-m-long crane boom. if u = 30, determine the placement x of the hook

Illusion [34]

Answer:

The answer is 80 kN . m (clockwise)

Explanation:

As,

M = P x L

Here, the towline exerts a force is P.

Substituting P for 4000N.

M = -4000N x 20m

= -80000N.m

= 80kN.m

Maximum moment about the point O is 80kN.m (Clockwise)

7 0

3 years ago

The drawing shows three identical springs hanging from the ceiling. Nothing is attached to the first spring, whereas a 4.5 n blo

Naya [18.7K]

<Continuation of the question>

(a) the spring constant (in N/m) and

(b) the weight of the block hanging from the third spring.

the distance for the first spring is 20cm, the second 35cm, the third 50cm.

Answer:

a) To find the spring constant, we'll use the formula

F=kx, if we make k the subject we'll get

k=F/x, where F=4.5N, x = 35cm - 20cm = 15cm = 0.15m

k=F/x = 4.5N/0.15m = 30N/m is the Answer

b) to find the weight of block hanging on third spring

we use the formula F=kx

where k = 30N/m, x=50cm-20cm=30cm=0.30m

F=kx = (30N/m)*(0.30m) = 9N is the Answer

4 0

3 years ago

Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular orbit on a Hohmann transfer traject

Finger [1]

Answer:

t = 12,105.96 sec

Explanation:

Given data:

weight of spacecraft is 2000 kg

circular orbit distance to saturn = 180 km

specific impulse = 300 sec

saturn orbit around the sun R_2 = 1.43 *10^9 km

earth orbit around the sun R_1= 149.6 * 10^ 6 km

time required for the mission is given as t

$t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}$

where

$\mu_{sun}$ is gravitational parameter of sun = 1.32712 x 10^20 m^3 s^2. $t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}$

t = 12,105.96 sec

6 0

3 years ago

A rigid rectangular loop, which measures 0.30 m by 0.40 m, carries a current of 5.5 A, as shown in the figure. A uniform externa

slamgirl [31]

Answer:

The magnitude will be " $(1.097 \ N/m)\hat{j}$ ". The further explanation is given below.

Explanation:

Trying to determine what is usual for a rectangular loop plane,

⇒ $\hat{n}=(Cos35^{\circ})(-\hat{i})+(Sin35^{\circ})(\hat{k})$

Magnetic moment is given as:

μ = $IA\hat{n}$

On putting the values in the above formula, we get

μ = $(5.5A)[(0.3m)(0.4m)][(Cos35^{\circ})(\hat{i})+(Sin35^{\circ})(-\hat{k})]$

= $0.66 Am^2[(Cos35^{\circ})(\hat{i})+(Sin35^{\circ})(\hat{k})]$

Now the external vector-shaped torque seems to be:

Magnitude,

$\sigma=\hat{\mu}\times\hat{\beta}$

On putting the values in the above formula, we get

⇒ $\sigma=[(0.06Am^2)(Cos35^{\circ}(-\hat{i})+Sin35^{\circ}(-\hat{k})]\times (2.9T)(-\hat{i})$

⇒ $= (0.66)(2.9)Cos35^{\circ}(\hat{i}\times \hat{i})+(0.66)(2.9)Sin35^{\circ}(\hat{k}\times \hat{k})$

⇒ $= 0+(1.97 \ N/m)\hat{j}$

⇒ $=(1.097 \ N/m)\hat{j}$

8 0

3 years ago

One __________________________ is the amount of energy needed to apply a force of 1 Newton over a distance of 1 meter.

lina2011 [118]

Answer:

Joule

Explanation:

I took the test

6 0

3 years ago