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2 years ago

13

I have a 3 cc piece of aluminum with a density of steel with a mass of 24.0 g. I cut it into 2 equal pieces. How has the density

of the steel pieces changed

1 answer:

tresset_1 [31]2 years ago

3 0

Answer:

What happens to the density of an object if the object is cut in half? ... The density remains the same because cutting the object in half will divide the mass & volume by the same amount. Also, the density of a substance remains the same no matter what size it is.

Explanation:

What happens to the density of an object if the object is cut in half? ... The density remains the same because cutting the object in half will divide the mass & volume by the same amount. Also, the density of a substance remains the same no matter what size it is.

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A student conducts an experiment in which a cart is pulled by a variable applied force during a 2 s time interval. In trial 1, t

fredd [130]

Answer:

The answer is "Including all three studies of 0s to 2s, that shift in momentum is equal".

Explanation:

Its shift in momentum doesn't really depend on the magnitude of its cars since the forces or time are similar throughout all vehicles.

Let's look at the speed of the car

$F = m a\\\\a =\frac{F}{m}$

We use movies and find lips

$\to v = v_0 + a t\\\\\to v = v_0 + (\frac{F}{m}) t$

The moment is defined by

$\to p = m v$

The moment change

$\Delta p = m v - m v_0$

Let's replace the speeds in this equation

$\Delta p = m (v_0 + \frac{F}{m t}) - m v_0\\\\\Delta p = m v_0 + F t - m v_0\\\\\Delta p = F t$

They see that shift is not directly proportional to the mass of cars since the force and time were the same across all cars.

5 0

2 years ago

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A cons

Virty [35]

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

$H_m=1.65m$

$H_E=1.16307m$

Explanation:

From the question we are told that

Mass of ball $M=2kg$

Length of string $L= 2m$

Wind force $F=13.2N$

Generally the equation for $\angle \theta$ is mathematically given as

$tan\theta=\frac{F}{mg}$

$\theta=tan^-^1\frac{F}{mg}$

$\theta=tan^-^1\frac{13.2}{2*2}$

$\theta=73.14\textdegree$

Max angle = $2*\theta= 2*73.14=>146.28\textdegree$

Generally the equation for max Height $H_m$ is mathematically given as

$H_m=L(1-cos146.28)$

$H_m=0.9(1+0.8318)$

$H_m=1.65m$

Generally the equation for Equilibrium Height $H_E$ is mathematically given as

$H_E=L(1-cos73.14)$

$H_E=0.9(1+0.2923)$

$H_E=1.16307m$

8 0

2 years ago

A car is brought to rest in a distance of 484m using a constant acceleration of -8.0m/s^2. What was the velocity of the car when

Agata [3.3K]

Answer:

88 m/s

Explanation:

To solve the problem, we can use the following SUVAT equation:

$v^2-u^2=2ad$

where

v is the final velocity

u is the initial velocity

a is the acceleration

d is the distance covered

For the car in this problem, we have

d = 484 m is the stopping distance

v = 0 is the final velocity

$a=-8.0 m/s^2$ is the acceleration

Solving for u, we find the initial velocity:

$u=\sqrt{v^2-2ad}=\sqrt{-2(8.0)(484)}=88 m/s$

6 0

2 years ago

A mountain-climber friend with a mass of 74 kg ponders the idea of attaching a helium-filled balloon to himself to effectively r

bazaltina [42]

Answer:

$V=16.65 m^3$

Explanation:

The volume of the balloon can be find compared the force in each cases so:

reduce 25% from 74kg

$R=\frac{25}{100}*74kg=18.5kg$

So the net force uproad on the balloon is

$F_b=18.5kg*g$

Now the density of the both gases air and helium are different however the volume is the same change offcorss the mass so:

$P_h=\frac{m}{V}=0.179 kg/m^3$

$P_A=1.29 kg/m^3$

$F_b=F_A-F_H$

$F_b=m_a*g-m_h*g$

$m=P/V$

$18.5kg*g=(1.29kg/m^3-0.179kg/m^3*)V*g$

$V=\frac{18.5kg}{(1.29-0.179)kg/m^3}$

$V=16.65 m^3$

4 0

2 years ago

Select all the correct answers.

sdas [7]

I believe there are two correct answers, and those answers are A and D

5 0

2 years ago