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2 years ago

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An auto moves 10 meters in the first second of travel, 15 more meters in the next second, and 20 more meters during the third se

cond. The acceleration of the auto is

1 answer:

k0ka [10]2 years ago

6 0

im guessing it's 5 m/s

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The acceleration due to gravity for any object, including 1 washer on the string, is always assumed to be m/s2. The mass of 3 wa

Pachacha [2.7K]

Answer:

The force will increase in proportion to the mass of the objects

Explanation:

The acceleration due to gravity is always the same. It is expressed in meters per second squared or m/s². The figure of 9.81 m/s² is an average value that was taken after calculating the acceleration under different surfaces. In fact, the acceleration differs depending on the shape of the part of the earth in relation to the earth's magnetic field and force.

Thus, if one washer was 20 kg, the acceleration being 9.81 m/s² the weight will be:

F = ma

= $20 * 9.81\\= 196.2 N$

If there are there washers, the weight will be:

F = 3 * 20 * 9.81

= 588.6 N

5 0

2 years ago

Read 2 more answers

An 18-cm-long bicycle crank arm, with a pedal at one end, is attached to a 20-cm-diameter sprocket, the toothed disk around whic

Tpy6a [65]

Answer:

Part a)

$a = 0.056 m/s^2$

Part b)

$L = 7.85 m$

Explanation:

Part a)

Angular speed of the pedal is changing from 60 rpm to 90 rpm in 10 s

so the angular acceleration is given as

$\alpha = \frac{\omega_2 - \omega_1}{\Delta t}$

so we will have

$\alpha = \frac{2\pi(\frac{90}{60}) - 2\pi(\frac{60}{60})}{10}$

$\alpha = 0.314 rad/s^2$

now the tangential acceleration of the pedal is given as

$a = r \alpha$

$a = 0.18 \times 0.314$

$a = 0.056 m/s^2$

Part b)

Total angular displacement made by the sprocket in the interval of 10 s is given as

$\theta = \frac{\omega_f + \omega_i}{2} t$

$\theta = \frac{2\pi (\frac{90}{60}) + 2\pi (\frac{60}{60})}{2}(10)$

$\theta = 78.5 radian$

now length of the chain passing over it is given as

$L = R\theta$

$L = 0.10 \times 78.5$

$L = 7.85 m$

6 0

2 years ago

a pebble is dropped down a well and hits the water 1.5 seconds later. using the equations for motion with constant acceleration,

Effectus [21]

Let h = the distance from the edge of the wall to the water surface (m).

Use g = 9.8 m/s² and neglect air resistance.

The initial vertical velocity of the pebble is zero.
Because the pebble hits the surface of the water after 1.5 s, therefore
h = (1/2)*(9.8 m/s²)*(1.5 s)² = 11.025 m

Answer: 11.025 m

7 0

2 years ago

Read 2 more answers

Suppose we replace the mass in the video with one that is four times heavier. How far from the free end must we place the pivot

Llana [10]

We must place the pivot to keep the meter stick in balance at 90 cm (10 cm from the weight) from the free end.

Answer: Option B

<u>Explanation:</u>

In initial stage, the meter stick’s mass and mass hanged in meter stick at one end are same. Refer figure 1, the mater stick’s weight acts at the stick’s mid-point.

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

$m \times g \times(x)+((m \times g)(x-50 \mathrm{cm}))=0$

$(m \times g \times x)-(50 \times m \times g)+(m \times g \times x)=0$

Taking out ‘mg’ as common and we get

$2 x-50=0$

$2 x=50$

$x=\frac{50}{2}=25 \mathrm{cm}$

Hence, the stick should be pivoted at a distance of,

$x^{\prime}=100 \mathrm{cm}-25 \mathrm{cm}=75 \mathrm{cm}$

So, the stick should be pivoted at a distance of 75 cm at the free end

Now, replace mass with another mass. i.e., four times the initial mass (as given)

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

$4 m g(x)+(m g)(x-50 c m)=0$

$4 m g x+m g x-50 m g=0$

Taking out ‘mg’ as common and we get

$5 x=50$

$x=\frac{50}{5}=10 \mathrm{cm}$

Hence, the stick should be pivoted at a distance of,

$x^{\prime}=100 \mathrm{cm}-10 \mathrm{cm}=10 \mathrm{cm}$

So, the stick should be pivoted at a distance of 10 cm from the free end.

Therefore, the option B is correct 90 cm (10 cm from the weight).

3 0

2 years ago

A small mass m is tied to a string of length L and is whirled in vertical circular motion. The speed of the mass v is such that

adell [148]

Answer:

(mv^2/R)/(mg)=1/2

v^2=R/2g

7 0

1 year ago