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3 years ago

Please help i don't understand. will mark brainliest

Physics

2 answers:

soldier1979 [14.2K]3 years ago

8 0

Answer:

marblebrainiest plz\c cmarble

Explanation:

dmitriy555 [2]3 years ago

4 0

Answer: It is A Soccer Ball

Explanation:

The soccer ball has more force than everything else!

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At what time was the person at a position of 0m?

Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

= (0 x 1) + 1/2 (0 x 1 x 1)

= (0) + 1/2 (0)

= 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

= 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1) = 10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0

3 years ago

En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a

Maru [420]

Answer:

c=0.14J/gC

Explanation:

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

We can use the expression for heat transmission

$Q=mc(T_2-T_1)$

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

$Q_1=-Q_2$

for water we have to

c = 4.18J / g ° C

replacing we have

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

I hope this is useful for you

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

Podemos usar la expresión para la transmisión de calor

$Q=mc(T_2-T_1)$

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

$Q_1=-Q_2$

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

7 0

3 years ago

While tuning a string to the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string.

lara31 [8.8K]

Answer:

a)the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b) 526hz

c)0.989 or a 1.14% decrease in tension

Explanation:

a) While tuning a string at 523 Hz,piano tuner hears 2.00 beats/s between a reference oscillator and the string.

The possible frequencies of the string can be calculated by

fl=f' - B

where

fl= lower limit of the possible frequency

f'= frequency of the string

B= beat heard by the tuner

fl= 523hz + Or - (2beats/secs * 1hz/1beat per sc)

fl= 521hz or 525hz

So the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b)fl=f' - B

523hz= f' - 3

f'= 523 + 3= 526hz

c) The tension is directly proportional to the square of the frequencies

T1/T2 =f1^2/f2^2

523^2 / 526^2 = 0.989 or a 1.14% decrease in tensio

6 0

3 years ago

A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi

slega [8]

We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
$dI =r^2*dm$
where $dm =M(b*dr)/(ab)$
Now we find the moment of inertia by integrating from $-a/2$ to $a/2$
The moment of inertia is
$I= \int\limits^{-a/2}_{a/2} {r^2*dm} = M \int\limits^{-a/2}_{a/2} r^2(b*dr)/(ab)=(M/a)(r^3/3)$ (from (-a/2) to $I=(M/3a)(a^3/8 +a^3/8)=(Ma^2)/12$ (a/2))

4 0

3 years ago

What is the relation between the gravitational force acting between any two bodies object with their masses and the distance bet

LenaWriter [7]

Answer:

The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. ... increases, the force of gravity decreases. If the distance is doubled, the force of gravity is one-fourth as strong as before.

7 0

3 years ago