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3 years ago

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Se lanza hacia arriba un objeto con una velocidad de 72 Dam/min y al mismo tiempo se deja caer una pelota desde una altura de 32

m. Calcula con las fórmulas de movimiento: a) La altura máxima alcanzada por el objeto b) La altura a la que se van a encontrar el objeto y la pelota c) La velocidad del objeto y de la pelota cuando lleguen al suelo

2 answers:

HACTEHA [7]3 years ago

8 0

Answer:

Q DIFICIL TIOO

guapka [62]3 years ago

5 0

Answde147

Explanation:

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You are designing a hydraulic lift for a machine shop. The average mass of a car it needs to lift is about 1500 kg. You wish to

ehidna [41]

Answer:

(a) Area(small piston)/Area(large piston) = 0.037

(b) h = 1336.36 cm = 13.36 m

Explanation:

(a)

The stress on the smaller piston is equally transmitted to the larger piston, in a hydraulic lift. Therefore,

Stress (small piston) = Stress (large piston)

Force (small piston)/Area (small piston) = Force (Large Piston)/Area (Large Piston)

Area(small piston)/Area(large piston) = Force (small piston)/Force(Large piston)

Area(small piston)/Area(large piston) = 550 N/(1500 kg)(9.8 m/s²)

<u>Area(small piston)/Area(large piston) = 0.037</u>

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(b)

The work is also transmitted equally to the large piston. So,

Work(small piston) = Work(Large Piston)

Force(small piston).Displacement(small piston) = Force(large piston).Displacement(small piston)

(550 N)(h) = (1500 kg)(9.8 m/s²)(50 cm)

h = 735000 N.cm/550 N

<u>h = 1336.36 cm = 13.36 m</u>

5 0

3 years ago

What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force $F_{1}=18\ N$

Force $F_{2}=26\ N$

Force $F_{3}=14\ N$

We need to calculate the torque due to force F₁

Using formula of torque

$\tau_{1}=-F_{1}d_{1}$

$\tau_{1}=-F_{1}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{1}=-18\times\dfrac{0.2}{2}$

$\tau_{1}=-1.8\ N-m$

We need to calculate the torque due to force F₂

Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

Hence, The net torque on the square plate is 2.72 N-m.

3 0

3 years ago

How often do the earth's magnetic poles switch?

RSB [31]

Reversals are the rule, not the exception. Earth has settled in the last 20 million years into a pattern of a pole<span> reversal about every 200,000 to 300,000 years, although it has been more than twice that </span>long<span> since the last reversal.</span>

3 0

3 years ago

Without effective assessment methods, teachers have no documented proof of childrens specific ____ and _____.

schepotkina [342]

Answer:

D strengths and weakneses

Explanation:

8 0

3 years ago

A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o

gulaghasi [49]

A. 0.77 A

Using the relationship:

$P=\frac{V^2}{R}$

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega$

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega$

The two light bulbs are connected in series, so their equivalent resistance is

$R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega$

The two light bulbs are connected to a voltage of

V = 240 V

So we can find the current through the two bulbs by using Ohm's law:

$I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A$

B. 142.3 W

The power dissipated in the first bulb is given by:

$P_1=I^2 R_1$

where

I = 0.77 A is the current

$R_1 = 240 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_1 = (0.77 A)^2 (240 \Omega)=142.3 W$

C. 42.7 W

The power dissipated in the second bulb is given by:

$P_2=I^2 R_2$

where

I = 0.77 A is the current

$R_2 = 72 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_2 = (0.77 A)^2 (72 \Omega)=42.7 W$

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

$P=\frac{E}{t}$

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0

2 years ago