Answer:
d. 6.0 m
Explanation:
Given;
initial velocity of the car, u = 7.0 m/s
distance traveled by the car, d = 1.5 m
Assuming the car to be decelerating at a constant rate when the brakes were applied;
v² = u² + 2(-a)s
v² = u² - 2as
where;
v is the final velocity of the car when it stops
0 = u² - 2as
2as = u²
a = u² / 2s
a = (7)² / (2 x 1.5)
a = 16.333 m/s
When the velocity is 14 m/s
v² = u² - 2as
0 = u² - 2as
2as = u²
s = u² / 2a
s = (14)² / (2 x 16.333)
s = 6.0 m
Therefore, If the car had been moving at 14 m/s, it would have traveled 6.0 m before stopping.
The correct option is d
Answer:
c=0.14J/gC
Explanation:
A.
2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.
B.
We can use the expression for heat transmission
In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say
for water we have to
c = 4.18J / g ° C
replacing we have
I hope this is useful for you
A.
2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.
B.
Podemos usar la expresión para la transmisión de calor
En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir
para el agua tenemos que
c=4.18J/g°C
reemplazando tenemos
The time taken for the light to travel from the camera to someone standing 7 m away is 2.33×10¯⁸ s
Speed is simply defined as the distance travelled per unit time. Mathematically, it is expressed as:
<h3>Speed = distance / time </h3>
With the above formula, we can obtain the time taken for the light to travel from the camera to someone standing 7 m away. This can be obtained as follow:
Distance = 7 m
Speed of light = 3×10⁸ m/s
<h3>Time =?</h3>
Time = Distance / speed
Time = 7 / 3×10⁸
<h3>Time = 2.33×10¯⁸ s</h3>
Therefore, the time taken for the light to travel from the camera to someone standing 7 m away is 2.33×10¯⁸ s
Learn more: brainly.com/question/14988345
Based on Newton's principle, whenever objects A and B interact with each other, they exert forces upon each other.
When a horse pulls on a cart, t<span>he horse exerts a force only to the cart. But that force applies only to the cart, not to the horse.
The cart in turn exerts a force on the horse. But that force applies only to the horse, not the cart also.
</span>
There are two forces resulting from this interaction - a force on the horse and a force on the cart. T<span>he net force on the cart remains as it was --- a positive force in the direction of the horse's movement. Therefore, the cart begins to accelerate and move.</span><span>
</span>
The correct answer is true