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3 years ago

1 - It is okay to use slang on a record as long as the auditor can interpret it.

1 answer:

6 0

1. false 2. false 3. true 4. not sure 5. b 6. b or d 7. not sure 8.not sure 9. not sure 10. c

lol sorry if i’m wrong on any i’m just using common sense

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A dragster starts with zero velocity and completes a 404.5 m (0.2528 mile) run in 4.922 s. If the car had a constant acceleratio

KIM [24]

Answer:

a. a=33.34ms⁻², V=164.4m/s

Explanation:

Since the dragster started with zero velocity, de determine the acceleration using of the equations of motion.

Below are the data given

Distance, s=404.5m,

time taken,t=4.922secs

Using the equation

S=ut+1/2at²

where u is the initial velocity and u=0

Making the acceleration the subject of the formula, we arrive at

a=2s/t²

a=(2*404.5)/4.922²

a=33.34ms⁻².

To determine the velocity, we use

V=u+at

V=0+33.34ms⁻² *4.922sec

V=164.4m/s

5 0

3 years ago

Read 2 more answers

How does groundwater move through the lithosphere?

Viktor [21]

Aquifer- the layer permeable rock that has connecting pores & transmit water freely, the artesian well water rises to the surface.

Hope this helps!

5 0

3 years ago

A 5.0 kg box is sliding across a waxed floor by the application of a 15 N east force. If the force of friction is 2.5 N west, wh

coldgirl [10]

1) 12.5 N east

There are two forces acting on the box along the horizontal direction:

- The applied force of 15 N east, we indicate it with F

- The force of friction of 2.5 N west, we indicate it with $F_f$

Taking east as positive direction, we can write the two forces has

$F=+15 N\\F_f = -2.5 N$

Therefore, the net force on the box will be:

$F_{net} = F + F_f = 15 + (-2.5) = +12.5 N$

And the positive sign means the direction is east.

2) $2.5 m/s^2$

We can solve this part by using Newton's second law:

$F_{net}=ma$

where

$F_{net}$ is the net force on the box

m is its mass

a is the acceleration

For the box in this problem,

$F_{net} = 12.5 m/s^2$ (east)

m = 5.0 kg

Solving for a, we find the acceleration:

$a=\frac{F_{net}}{m}=\frac{12.5}{5.0}=2.5 m/s^2$

And the direction is the same as the net force (east)

6 0

4 years ago

Imagine holding a basketball in both hands, throwing it straight up as high as you can, and then catching it when it falls. At w

valentina_108 [34]

Answer:before throwing and after catching the ball

Explanation:

When basketball is in the hand of player net force on it zero as holding force is canceled by gravity Force. During its entire motion gravitational force is acting on the ball which is acting downward. Even at highest point gravity is constantly acting downwards.

After catching the ball net force on it zero as holding force is canceled by gravity force and ball is continue to be in stationary motion.

6 0

3 years ago

Una bola de 1 kg gira alrededor de un circulovrtical en el extremo de un cuerda. El otro extremo de la cuerda esta fijo en el ce

Vladimir79 [104]

Answer:

La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

Explanation:

Puesto que la bola gira en un círculo vertical, existe claramente una diferencia entre las tensiones debido a la influencia de la gravedad y la tensión que resulta de la aceleración centrípeta experimentada por la masa. La máxima tensión ocurre cuando la bola se encuentra en el nadir (o la sima) del trayecto circular, la cual se describe por la Segunda Ley de Newton:

$T_{max} - m\cdot g = m\cdot \frac{v^{2}}{L}$