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Salsk061 [2.6K]
3 years ago
10

Based on the data given, in what direction will the car accelerate?

Physics
1 answer:
skad [1K]3 years ago
7 0
Vertical forces:
There is a force of 579N acting upward, and a force of 579N
acting downward.
The vertical forces are balanced ... they add up to zero ...
so there's no vertical acceleration. 
Not up, not down.

Horizontal forces:
There is a force of 487N acting to the left, and a force of 632N
acting to the right.
The net horizontal force is

        (487-left + 632-right)  -  (632-right - 487-right) =  145N to the right.

The net force on the car is all to the right.
The car accelerates to the right.
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The position of the sun and the moon affect how high the tide is 
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A flea jumps straight up to a maximum height of 0.540 m . what is its initial velocity v0 as it leaves the ground?
jok3333 [9.3K]
For an uniformly accelerated motion, we can write
2ah=v_f^2-v_0^2
where a=g=-9.81 m/s^2 is the acceleration of this motion, which in this problem is the gravitational acceleration, with a negative sign because it points downward, against the direction of the motion; h=0.540 m is the distance covered by the flea, and v_0 is the initial velocity. 

At the maximum height, the velocity is zero, so v_f =0. Therefore we  can solve to find v_0:
v_0 =  \sqrt{2ah}= \sqrt{2(9.81 m/s^2)(0.540 m)}  =3.25 m/s
3 0
3 years ago
Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. one force is 9.0 n, acting due east, an
Zolol [24]
The first thing you should do for this case is to find the horizontal and vertical components of the forces acting on the body.
 We have then:
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 Then, the resulting net force is:
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6 0
3 years ago
The answer is -7m, but I don’t know how that’s the answer. Can anyone explain?
atroni [7]

Explanation:

At first it is in 14m position but position doesn't matter in displacement, similar case for time taken.

So at first it travels 6m in positive direction.

So displacement= 6m

Then it travels 13 in opposite or negative direction.

So displacement = 6 -13 m = -7 m

Hope it helps ya

8 0
3 years ago
Read 2 more answers
2.- Si una cámara fotográfica emite un pulso de sonido para enfocar un objeto, determinar
uranmaximum [27]

Answer:

a. El tiempo de recorrido es 5.882\times 10^{-3} segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

b. El tiempo de recorrido es 0.118 segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

Explanation:

El sonido es un tipo de onda mecánica, que es un tipo de onda que necesita de un medio material para propagarse. En este caso, entendemos que el sonido se propaga a través del aire atmosférico hasta llegar a su destino y devolverse a rapidez constante. Entonces, podemos estimar el tiempo (t), medido en segundos, a partir de la siguiente fórmula:

t = \frac{2\cdot x_{s}}{v_{s}}

Donde:

x_{s} - Distancia entre la cámara fotográfica y el objeto, medida en metros.

v_{s} - Rapidez del sonido en el aire atmosférico, medida en metros por segundo.

A continuación, calculamos el tiempo de recorrido:

a. (x_{s} = 1\,m, v_{s} = 340\,\frac{m}{s})

t = \frac{2\cdot (1\,m)}{340\,\frac{m}{s} }

t = 5.882\times 10^{-3}\,s

El tiempo de recorrido es 5.882\times 10^{-3} segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

b. (x_{s} = 20\,m, v_{s} = 340\,\frac{m}{s})

t = \frac{2\cdot (20\,m)}{340\,\frac{m}{s} }

t = 0.118\,s

El tiempo de recorrido es 0.118 segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

8 0
3 years ago
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