Which shows one example of physical change and the one example of a chemical change?

I know the acceleration due to gravity (ie 9.8 m/s2) will have a negative sign when falling down, a positive one when going up.

rewona [7]

Answer:

Yes

Explanation:

Accerelation is measured by change in velocity. So naturally, if an object is slowing down, its velocity is decreasing so acceleration is negative. If it is speeding up velocity is increasing so positive acceleration.

(Velocity final - Velocity initial)/t

Note that this does not apply only to gravity, but to all linear accelerations

6 0

2 years ago

As a rocket ascends, its acceleration increases even though the net force on it stays constant. why? (assume a traveling distanc

topjm [15]

<span>the rocket's mass decreases as its fuel is consumed. the same net force acting on a smaller mass results in a larger acceleration</span>

5 0

3 years ago

Read 2 more answers

An arrow is shot at 27.0° above the horizontal. Its velocity is 54 m/s, and it hits the target.

GuDViN [60]

Explanation:

(a) What is the maximum height the arrow will attain?

Given:

v₀ᵧ = 54 sin 27.0° m/s = 24.5 m/s

vᵧ = 0 m/s

aᵧ = -9.8 m/s²

Find: Δy

vᵧ² = v₀ᵧ² + 2aᵧΔy

(0 m/s)² = (24.5 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 30.7 m

(b) The target is at the height from which the arrow was shot. How far away is it?

Given:

Δy = 0 m

v₀ᵧ = 54 sin 27.0° m/s = 24.5 m/s

aᵧ = -9.8 m/s²

Find: t

Δy = v₀ᵧ t + ½ aᵧt²

0 m = (24.5 m/s) t + ½ (-9.8 m/s²) t²

t = 5.00 s

Given:

v₀ₓ = 54 cos 27.0° m/s = 48.1 m/s

aₓ = 0 m/s²

t = 5.00 s

Find: Δx

Δx = v₀ₓ t + ½ aₓt²

Δx = (48.1 m/s) (5.00 s) + ½ (0 m/s²) (5.00 s)²

Δx = 241 m

3 0

3 years ago

calculate the density of a neutron star with a radius 1.05 x10^4 m, assuming the mass is distributed uniformly. Treat the neutro

Orlov [11]

To develop this problem it is necessary to apply the concepts related to the proportion of a neutron star referring to the sun and density as a function of mass and volume.

Mathematically it can be expressed as

$\rho = \frac{m}{V}$

Where

m = Mass (Neutron at this case)

V = Volume

The mass of the neutron star is 1.4times to that of the mass of the sun

The volume of a sphere is determined by the equation

$V = \frac{4}{3}\pi R^3$

Replacing at the equation we have that

$\rho = \frac{1.4m_{sun}}{\frac{4}{3}\pi R^3}$

$\rho = \frac{1.4(1.989*10^{30})}{\frac{4}{3}\pi (1.05*10^4)^3}$

$\rho = 5.75*10^{17}kg/m^3$

Therefore the density of a neutron star is $5.75*10^{17}kg/m^3$

4 0

3 years ago

A string or rope will break apart if it is placed under too much tensile stress. Thicker ropes can withstand more tension withou

Lemur [1.5K]

Answer: The answer is 13.02 meter

Explanation: your question is incomplete, the complete question is this if A string or rope will break apart if it is placed under too much tensile stress. Thicker ropes can withstand more tension without breaking because the thicker the rope, the greater the cross-sectional area and the smaller the stress. One type of steel has a density of

7750kg/m3 and will break if the tensile stress exceeds 7.0×108N/m2.

You want to make a guitar string from a mass of 4.5 g of this type of steel. In use, the guitar string must be able to withstand a tension of 900 N without breaking. Your job is the following.

(a) Determine the maximum length the string can have

Answer:

As we know that density(p)= mass(m)/volume(V)

Solve for volume(V)= mass/density

If the given mass value is in gram then we have to divide it by 1000 to get the value in kilogram so our mass in kg will be 3.7/1000= 0.0037 kg

And if density is in kg/m^3

Then volume will be V= 0.0037/7730 =>

4.79^10-7m^3

Now we know that stress = Force/Area

For Area = force/stress => 900/7*10^8=> 2795.95m^2

For length= volume/Area => 4.79^10-7/2795.95

Length= 13.02 m

4 0

3 years ago