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3 years ago

ANYONE PLEASE HELP

2 answers:

8 0

<span>C.) Atoms that are arranged differently create different substances with different properties

It can't be A, 'cause both have Covalent bonds, plus it can't be B, 'cause their properties are not same/

Hope this helps!</span>

Setler79 [48]3 years ago

7 0

The answer is C

Giddy up

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If a lamp in your house starts flickering, and you notice that the cord has a frayed wire, then the lamp has probably developed

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It has probably developed an Short Circuit.

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4 years ago

slader How much energy is required to move a 1040 kg object from the Earth's surface to an altitude four times the Earth's radiu

andrew-mc [135]

Answer:

ΔU = 5.21 × 10^(10) J

Explanation:

We are given;

Mass of object; m = 1040 kg

To solve this, we will use the formula for potential energy which is;

U = -GMm/r

But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.

Thus;

ΔU = -GMm((1/r_f) - (1/r_i))

Where;

M is mass of earth = 5.98 × 10^(24) kg

r_f is final radius

r_i is initial radius

G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Since, it's moving to altitude four times the Earth's radius, it means that;

r_i = R_e

r_f = R_e + 4R_e = 5R_e

Where R_e is radius of earth = 6371 × 10³ m

Thus;

ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)

× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))

ΔU = 5.21 × 10^(10) J

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3 years ago

Two spheres are made of wood - the first is a variety of wood whose density is equal to that of water, while the second is of a

ki77a [65]

Answer:

The buoyant force experienced by a body is equal to product of unit weight of liguid in which the the objevt is immersed and the volume of liquid replaced by the object.

In the given scenario, bothe the spheres have equal volume and are fully submerged in water. Therefore, the buoyant force experienced by both the spheres will be equal.

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4 years ago

A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.65 m/s2 for 17.0 s. 2

nlexa [21]

Hello

Let's solve the problem in the three different steps

1) Uniformly accelerated motion, with acceleration $a_1 = 2.65~m/s^2$ and for a total time of $t_1=17~s$ . The body is initially at rest, so the distance covered is given by
$S= \frac{1}{2}a_1t_1^2=382.9~m$
Calling $v_f$ and $v_i$ the final and initial velocity, and since the $v_i=0~m/s$ because the body starts from rest, we can use
$a= \frac{v_f-v_i}{t}$
to find the final velocity after this first leg:
$v_{f}=v_i+a_1t_1=45~m/s$
And the average velocity in this first leg is
$v_1= \frac{v_f+v_i}{2}=22.5~m/s$

2) Uniform motion. The velocity is constant and it is equal to the final velocity of the first leg: $v_2=45~m/s$ . This is also the average velocity of the second leg.
The total time of this second leg is $t_2=1.60~min = 96~s$ . The distance covered is given by
$S_2=v_2t_2=45~m/s \cdot 96~s=4320~m$

3) Uniformly decelerated motion, with constant deceleration of $a_3=-9.39~m/s^2$ and for a total time of $t_3=4.8~s$ . Here, the initial velocity of the body is the final velocity of the previous leg, i.e. $v_i=45~m/s$ . Therefore, the distance covered in this leg is given by
$S_3=v_i t_3 + \frac{1}{2} a_3 t^2 =107.8~m$
The final velocity in this leg is given by
$v_f=v_i+at=45~m/s-9.39~m/s^2 \cdot 4.8~s = -0.07~m/s$
The negative sign means that after decelerating, the body has started to go in the opposite direction. Similarly to step 1, the average velocity in this leg is given by
$v_3 = \frac{1}{2}(v_f+v_i)= \frac{1}{2}(-0.07~m/s+45~m/s)= 22.5~m/s$

4) Finally, the total distance covered in the motion is
$S=S_1+S_2+S_3=382.9~m+4320~m+107.8~m=4810.7~m$
To find the average velocity, we must "weigh" the average velocity of each leg for the correspondent time of that leg:
$v_{ave}= \frac{v_1t_1+v_2t_2+v_3t_3}{t_1+t_2+t_3}=40.8~m/s$

8 0

3 years ago

During heavy rain, a section of a mountainside measuring 2.3 km horizontally (perpendicular to the slope), 0.75 km up along the

Jobisdone [24]

Answer:

93328 kg

Explanation:

$\rho$ = Density of mud = 1900 kg/m³