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Juli2301 [7.4K]

3 years ago

14

Thor deja caer su martillo de la torre Torre Stark que tiene 95 pisos, si al llegar el Mjolnir al piso han transcurrido 9.5 segu

ndos. Determina: a) La velocidad con la que el Mjonir golpea el suelo

1 answer:

NARA [144]3 years ago

6 0

Answer:

La magnitud de la velocidad con la que Mjonir golpea el suelo es 93.2 m/s.

Explanation:

Podemos encontrar la velocidad con la que el martillo golpea el suelo usando la siguiente ecuación:

$v_{f} = v_{0} - gt$ (1)

En donde:

$v_{f}$ : es la velocidad final =?

$v_{0}$ : es la velocidad inicial = 0 (lo deja caer)

g: es la gravedad = 9.81 m/s²

t: es el tiempo = 9.5 s

Al reemplazar los valores anteriores en la ecuación (1), tenemos:

$v_{f} = 0 - 9.81 m/s^{2}*9.5 s = -93.2 m/s$

El signo negativo se debe a que el vector velocidad está dirigido hacia abajo en el eje vertical (y).

Por lo tanto, la magnitud de la velocidad con la que Mjonir golpea el suelo es 93.2 m/s.

Espero que te sea de utilidad!

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two resistors of 20 ohm each are connected in a parallel with a battery of 10V. The total current passing through circuit is

Varvara68 [4.7K]

Answer:

1 Ampere.

Explanation:

From the question given above, the following data were obtained:

Resistor 1 (R₁) = 20 ohm

Resistor (R₂) = 20 ohm

Voltage (V) = 10 V

Current (I) =?

Next, we shall determine the equivalent resistance in the circuit. This can be obtained as follow:

Resistor 1 (R₁) = 20 ohm

Resistor (R₂) = 20 ohm

Equivalent Resistance (R) =?

Since the resistors are in parallel connection, the equivalent resistance can be obtained as follow:

R = (R₁ × R₂) / (R₁ + R₂)

R = (20 × 20) / (20 + 20)

R = 400 / 40

R = 10 ohm

Finally, we shall determine the total current in the circuit. This can be obtained as illustrated below:

Voltage (V) = 10 V

Equivalent Resistance (R) = 10 ohm

Current (I) =?

V = IR

10 = I × 10

Divide both side by 10

I = 10 / 10

I = 1 Ampere

Therefore, the total current in the circuit is 1 Ampere.

3 0

2 years ago

The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the

Lynna [10]

Part a

Answer: NO

We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.

Using the second equation of motion:

$s=ut+0.5at^2$

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.

It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, $a=-10.0 m/s^2$

$\Rightarrow s=23.9 m/s \times 0.75s+0.5\times (-10.0 m/s^2)\times (0.75 s)^2=15.11 m$

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.

Part b

Answer: 29.6 m/s

The maximum distance that car can travel is $s=44.0 m$

The acceleration is same, $a=-10.0 m/s^2$

The final velocity, v=0

Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

$v^2-u^2=2as$

$0-u^2=-2 \time 10.0m/s^2 \times 44.0 m\Rightarrow u=\sqrt{880 m^2/s^2}=29.6 m/s$

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.

7 0

3 years ago

TRUE OR FALSE? We can see every member of the electromagnetic spectrum?

Svetradugi [14.3K]

It is false. We see only a few members of <span>electromagnetic spectrum</span>

Hope it helps!

4 0

3 years ago

Read 2 more answers

I will be so thankful if u answer correctly!!

polet [3.4K]

(C) 200 N

Explanation:

The acceleration due to gravity on earth $g_{E}$ is given by

$g_{E} = G\dfrac{M_{E}}{R_{E}^{2}}$

where $G$ = universal gravitational constant

$\:\:\:\:\:\:\:\:\:\:\:\:M_{R}$ = mass of the earth

$\:\:\:\:\:\:\:\:\:\:\:\:R_{E}$ = radius of the earth

Planet Krypton has twice the mass of earth and 3 times the radius so its acceleration due to gravity $g_{K}$ is

$g_{K} = G\dfrac{M_{K}}{R_{K}^{2}}$

$\:\:\:\:\:\: = G\dfrac{(2M_{E})}{(3R_{E})^{2}}$

$\:\:\:\:\;\:= (\dfrac{2}{9})G\dfrac{M_{E}}{R_{E}^{2}}$

or

$g_{K} = (\dfrac{2}{9})\:g_{E}$

If we multiply both sides by Superman's mass, we get his weights on both planets:

$mg_{K} = (\dfrac{2}{9})\:(mg_{E})$

$W_{K} = (\dfrac{2}{9})\:W_{E} = (\dfrac{2}{9})(900\:N)= 200\:N$

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3 years ago

A ford F-250 pulls a 1000-kg car with a net force of 2000 N. What is the acceleration of the car A=F/m

vfiekz [6]

Answer:

2ms^2

Explanation:

A = 2000 divided by 1000

A= 2

5 0

3 years ago