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3 years ago

if an object is moving with a velocity of 24m/s and has an acceleration of -4m/s how long will it take to stop

1 answer:

7 0

The negative sign on the acceleration is only a vector quantity that means the object is accelerating to the left. Hence, we can only focus on it magnitude which is 4 m/s^2. Acceleration is the change in velocity over time. The change in velocity must be 24 m/s - 0 m/s, if you want the object to stop. Therefore,

a = (v2 - v1)/t
4 = (24 - 0)t
t = 6 seconds

The object will stop after 6 seconds.

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What kind of force can be created with an electrical current?

RideAnS [48]

Answer:

magnatic force can be created

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3 years ago

What is the mass of a bird if it has a potential energy of 147J when flying at a height of 30 m above the ground?

Artist 52 [7]

Answer:

0.5kg

Explanation:

Given parameters:

Potential energy = 147J

Height = 30m

Unknown:

Mass of the bird = ?

Solution:

Potential energy is the energy due to the position of a body. Now, the expression for finding the potential energy is given as;

P.E = mgH

m is the mass

g is the acceleration due to gravity = 9.8m/s²

H is the height

147 = m x 9.8 x 30

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3 years ago

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

The radial acceleration is $a_c = 0.9574 m/s^2$

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

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3 years ago

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HACTEHA [7]

On behalf of the state, write an invitation for the siblings for the gala dinner

Hello there, we would like to welcome you to the gala dinner hope you can come :)

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3 years ago

During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0

oksian1 [2.3K]

Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

$d=\frac{1,280,000,000km}{37,000}=34,594.59km$

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

$v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}$

hence, the speed of the Hubble is approximately 384km/min

5 0

3 years ago