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babymother [125]
3 years ago
10

Silvia is driving at the speed limit of 25m/s when she sees an ambulance approaching in her rearview mirror. To move out of its

way, she accelerates to 32 m/s in 3 seconds. What was her acceleration.
Physics
1 answer:
lesya692 [45]3 years ago
6 0

a = ∆v/t  = (32 - 25) / 3 = 7/3 m/s2

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Debido al desorden en el laboratorio un científico tiene 2 termómetros diferentes pero no sabe en qué escalas están por lo que d
just olya [345]

Answer:

La escala del termómetro ''A'' es grados Celsius.

La escala del termómetro ''B'' es grados Fahrenheit.

Explanation:

Para hallar en qué escalas están los termómetros partimos de que la mezcla a la cuál se midió su temperatura mantuvo su temperatura constante.

Esto quiere decir que los termómetros están expresando la misma temperatura pero en una escala distinta.

Sabemos que dada una temperatura en grados Celsius ''C'' si la queremos convertir a grados Fahrenheit ''F'' debemos utilizar la siguiente ecuación :

F=(\frac{9}{5})C+32 (I)

Ahora, si reemplazamos y asumimos que la temperatura de 18° es en grados Celsius, entonces si reemplazamos C=18 en la ecuación (I) deberíamos obtener F=64.4 ⇒

F=(\frac{9}{5}).(18)+32=32.4+32=64.4

Efectivamente obtenemos el valor esperado. Finalmente, corroboramos que la temperatura del termómetro ''A'' está medida en grados Celsius y la temperatura del termómetro ''B'' en grados Fahrenheit.

6 0
2 years ago
I NEED MAJOR HELP ON THIS AS WELL PLEASE SOMEONE HELP ME
yuradex [85]

Answer:

The total distance is  130.2 [m]

Explanation:

In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the cart starts from rest, i.e. its initial speed is zero.

v_{f} =v_{o} +(a*t)

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

a = acceleration = 3 [m/s²]

t = time = 8[s]

Vf = 0 + (3*8)

Vf = 24 [m/s]

With this velocity we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

where

x = distance traveled [m]

24² = 0 + (2*3*x)

x = 576/(6)

x = 96 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

v_{f}= v_{i} -(a*t)

Vf = final velocity [m/s]

Vi = initial velocity = 24 [m/s]

a = desacceleration = 1.6 [m/s²]

t = time = 15 [s]

Vf = 24 - (1.6*15)

Vf = 21.6 [m/s]

With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} -2*a*x

where

x = distance traveled [m]

21.6² = 24² - (2*1.6*x)

x = 109.44/(3.2)

x = 34.2 [m]

Note: The negative sign in the equations is because the car is desaccelerating, it means its velocity is decreasing.

Therefore the total distance is Xt = 34.2 + 96 = 130.2 [m].

5 0
3 years ago
two cars start at the same point and drive in a straight line for 5km. At the end of the drive their distances are the same but
Anna11 [10]

A 'displacement' always consists of a magnitude and a direction.  The two cars you just described have displacements with the same magnitude ... 5 km.  But if they didn't both drive in the same direction, then their displacements are different.

Remember:

-- 10 m/s² up and 10 m/s² down are different accelerations

-- 30 mph East and 30 mph West are the same speed but different velocity.

-- 5 km North and 5 km South are the same distance but different displacement.

7 0
2 years ago
Could anyone help with number 9?
Alisiya [41]
The answer would be A
3 0
3 years ago
The linear charge density on the inner conductor is and the linear charge density on the outer conductor is
Lubov Fominskaja [6]

Complete Question

The complete question is  shown on the first uploaded image (reference for Photobucket )

Answer:

The  electric field is  E = -1.3 *10^{-4} \ N/C

Explanation:

 From the question we are told that

    The linear charge density on the inner conductor is  \lambda _i  =  -26.8 nC/m  =  -26.8 *10^{-9} C/m

    The linear charge density on the outer conductor is

  \lambda_o  = -60.0 nC/m  =  -60.0 *10^{-9} \ C/m

     The position of interest is r =  37.3 mm =0.0373 m

Now this position we are considering is within the outer conductor so the electric field  at this point is due to the inner conductor (This is because the charges on the conductor a taken to be on the surface of the conductor according to Gauss Law )

Generally according to Gauss Law

         E (2 \pi r l) =  \frac{ \lambda_i }{\epsilon_o}

=>    E =  \frac{\lambda _i }{2 \pi *  \epsilon_o * r}

substituting values  

       E =  \frac{ -26 *10^{-9} }{2 * 3.142 *  8.85 *10^{-12} * 0.0373}

       E = -1.3 *10^{-4} \ N/C

The  negative  sign tell us that the direction of the electric field is radially inwards

    =>   |E| = 1.3 *10^{-4} \ N/C

5 0
3 years ago
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