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scoundrel [369]

3 years ago

6

One type of ride falls straight down for 3 seconds. During this time, the ride accelerates from a speed of 0 m/s to a speed of 3

0 m/s. What is
the average acceleration of the ride?

Initial speed =

Final speed =

Time =

Average acceleration =

1 answer:

Alja [10]3 years ago

7 0

Answer:

Initial speed: 0m/s Final speed: 30m/s Time: 3 seconds Average Acceleration: 10m/s

Explanation:

I'm not sure if this is right... but you take the final speed and divide it by the time... just dont take advice from me anymore i guess

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A proton experiences a force of 4 Newton when it enters perpendicular to the direction of the magnetic field with a speed 100 m/

____ [38]

Answer:

F = 0 N

Explanation:

Force on a moving charge in constant magnetic field is given by the formula

$F = q(\vec v\times \vec B)$

so here it depends on the speed of charge, magnetic field and the angle between velocity of charge and the magnetic field

here when charge is moving with speed 100 m/s in a given magnetic field then the force on the charge is given as

$F = 4 N$

now when charge is moving parallel to the magnetic field with different speed then in that case

$\vec v \time \vec B = 0$

so here we have

F = 0

7 0

3 years ago

A small grinding wheel has a moment of inertia of 4.0*10-5kgm2. What net torque must be applied to the wheel for its angular acc

kvv77 [185]

Hi there!

We can use the rotational equivalent of Newton's Second Law:

$\huge\boxed{\Sigma \tau = I \alpha}$

Στ = Net Torque (Nm)

I = Moment of inertia (kgm²)

α = Angular acceleration (rad/sec²)

We can plug in the given values to solve.

$\Sigma \tau = (4 * 10^{-5})(150) = \boxed{0.006 Nm}$

4 0

3 years ago

A chemical is always the same thing as an element .<br><br> True <br> False

olya-2409 [2.1K]

That would be false

Hope this helps :)

3 0

3 years ago

Read 2 more answers

A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the sa

aleksandr82 [10.1K]

Answer: The value of the celsius temperature of the cube is 472.2°c.

Explanation:

The expression for the power radiated is as follows;

$P=A\epsilon\sigma T^{4}$

Here, A is the area, $\sigma$ is the stefan's constant, $\epsilon$ is the emissivity and T is the temperature.

It is given in the problem that A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the same emissivity as the sphere.

Then the expression for the radiated power for the cube and the sphere can be expressed as;

$A_{1}\epsilon \e\sigma T_{1}^{4}=A_{2}\epsilon \e\sigma T_{2}^{4}$

Here, $A_{1}$ is the area of the sphere, $A_{2}$ is the area of the cube, $T_{1}$ is the temperature of the sphere and $T_{2}$ is the temperature of the cube.

The radiated powers and emissivity of the cube and the sphere are same.

$A_{1}T_{1}^{4}=A_{2}T_{2}^{4}$

The area of the sphere is $A_{1}=4\pi \times r^{2}$ .

Here, r is the radius of the sphere.

The area of the cube is $A_{2}=6\times a^{2}$ .

Here, a is the edge of the cube.

Put $A_{1}=4\pi \times r^{2}$ and $A_{2}=6\times a^{2}$ .

$T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}$ ....(1)

The masses and the densities of the sphere and the cube are same. Then the volumes are also same.

$V_{1}=V_{2}$

Here, $V_{1},V_{1}$ are the volumes of the sphere and the cube.

$\frac{4}{3}\pi r^{3}=a^{3}$

$\frac{r}{a}=(\frac{3}{4\pi })^{\frac{1}{3}}$

Put this value in the equation (1).

$T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}$ $T_{2}=T_{1}(\frac{2\pi }{3}\times ((\frac{3}{4\pi })^{\frac{1}{3}})^{2})^{\frac{1}{4}}$

Put T_{1}=500°c.

$T_{2}=(500)(\frac{2\pi }{3}\times (\frac{3}{4\pi })^{\frac{2}{3}})^{\frac{1}{4}}$

$T_{2}=472.2^{\circ}c$

Therefore, the value of the celsius temperature of the cube is 472.7°c.

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3 years ago

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