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3 years ago

A book sitting on a table is moved horizontally. Describe the

Physics

1 answer:

Akimi4 [234]3 years ago

3 0

Frictional force and Applied force has same “magnitude” and “opposite” direction.

Option: B

<u>Explanation</u>:

When a book is moved horizontally by applying “force” on the book, the frictional force is opposed to the book by the table. Here, this “frictional force” is opposing the book has the same force what we applied on the book but this frictional force and the applied force are opposite in direction. Always the “frictional force” is opposite to the “applied force” which stops the object to move. For example, if a force applied leftward to the object the frictional force is acted on the right side of the object.

When two objects are in contact they experience a "frictional force". This "frictional force" acts opposite to the force applied on to move the object.

Formula for "frictional force" is $\mu\times N$

Where, $\mu$ is coefficient of friction and N is normal force.

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A racing car whose mass is 1.2 X 10^3 kg is travelling at 8.9 m/s. It stops with a constant deceleration in a distance of 1.8X10

Alexeev081 [22]

given that initial speed of the car is

$v_i = 8.9 m/s$

now after travelling the distance d = 1.8 * 10^1 m the car will stop

so here we can use kinematics to find the acceleration of car

$v_f^2 - v_i^2 = 2 a d$

$0 - 8.9^2 = 2 a d$

here we have

$- 79.21 = 2*(18)*a$

$a = -2.2 m/s^2$

net force applied due to brakes of car is given by Newton's II law

$F = ma$

here we have

mass = 1.2 * 10^3 kg

$F_{net} = 1.2 * 10^3 * 2.2$

$F_{net} = 2.64 * 10^3 N$

now we can say

$F_{net} = F_1 + F_2$

$2.64 * 10^3 = 1.8 * 10^3 + F_2$

$F_2 = 8.4 * 10^2 N$

So the force applied due to brakes is given as above

7 0

3 years ago

Einstein's theory of general relativity is currently the best explanation of gravity.Why has this theory not been replaced with

alina1380 [7]

Answer:

Option C, It still explains the experimental evidence pertaining to gravity

Explanation:

Please find the attachment

8 0

3 years ago

A bar of iron is 10 cm at 20°C. At 19°C it will be (are = 11 x 10-6/°C)

dangina [55]

Answer:

HENCE (3) IS ANSWER

Explanation:

α =ΔL/(LO×Δ T) or ΔL= α(LO×Δ T)

HERE ΔL IS CHANGE IN LENGTH, Lo is original length , α is coefficient of linear expansion , Δ T is change in temperature

α= 11 x 10 ^-6/°C , Δ T = 19-20 = -1°C

hence

ΔL= 11 x 10 ^-6×10×-1

= - 11×10^-5

MINUS SIGN SHOWS SHORTER

HENCE (3) IS ANSWER

4 0

3 years ago

Read 2 more answers

Water flows along a streamline down a river of constant width. Over a short distance, the water slows from speed v to v/3. Which

kvasek [131]

Answer:

a. It became deeper by a factor of 3.

Explanation:

What we have is water flowing down a river with constant width. The water slows from speed v to v3 over a shirt distance

Using the equation of continuity

A1V1 = A2V2 ----1

A1 is the area of rectangle

V1 is the velocity of water

Area of rectangle = length x width

We rewrite equation 1 as

λ1w1v1 = λ2w2v2

We have w1 = w2

λ1v1 = λ2v2

λ1*v1 = λ2*v/3

λ1 = λ2/3

So it becomes deeper by a factor of 3

8 0

3 years ago

Determine la resistencia equivalente de la "escalera" de

defon

Las respuestas a cada inciso son:

a) La resistencia equivalente de la "escalera" de resistores iguales de 125 Ω que se muestra en la figura adjunta es:

$R_{t} = 341.7 \: \Omega$

b) La corriente a través de cada uno de los tres resistores de la izquierda si se conecta una batería de 50.0 V entre los puntos A y B es:

Correspondiente a R8 y R9 es 0.23 A
Correspondiente a R7 es 0.17 A.

a) En la imagen adjuntada correspondiente a la Figura 26-40, podemos observar que las resistencias 1, 2 y 3 están en serie, por lo tanto la ressitencia equivalente entre estas 3 es:

$R' = R_{1} + R_{2} + R_{3}$

De aquí en adelante tendremos presente que las todas las resistencias son iguales entre sí y por ende igual a 125 Ω. Las notaciones del 1 al 9 son para poder mostrar la resolución del problema.

Entonces:

$R' = 3R$

Ahora, esta resistencia está en paralelo con la resistencia R₄, por lo tanto la resistencia equivalente entre estas dos es:

$\frac{1}{R''} = \frac{1}{R'} + \frac{1}{R_{4}} = \frac{1}{3R} + \frac{1}{R} = \frac{4R}{3R^{2}}$

$R'' = \frac{3}{4}R$

Luego, esta resistencia está en serie con las resistencias R₅ y R₆, por lo tanto:

$R''' = R'' + R_{5} + R_{6} = \frac{3}{4}R + 2R = \frac{11}{4}R$

Esta resistencia está ahora en paralelo con R₇, entonces:

$\frac{1}{R''''} = \frac{1}{R'''} + \frac{1}{R_{7}} = \frac{4}{11R} + \frac{1}{R} = \frac{15R}{11R^{2}}$

$R'''' = \frac{11}{15}R$

Finalmente, esta resistencis está en serie con las resistencias R₈ y R₉, por lo tanto la resistencia total es:

$R_{t} = R'''' + R_{8} + R_{9} = \frac{11}{15}R + 2R = \frac{41}{15}R = \frac{41}{15}*125 \: \Omega = 341.7 \: \Omega$

b) Para este inciso debemos usar la Ley de Kirchhoff, pues tenemos tres mallas. Supondremos que las corriente de cada malla fluiran en sentido horario, por lo tanto las ecuaciones de para cada malla serán:

Malla 1

Segun la ley de Ohm tenemos:

$V-i_{1}R-i_{3}R=0$ (1)

Malla 2

$-i_{2}R-i_{5}R-i_{2}R+i_{3}R=0$ (2)

Malla 3

$-i_{4}R-i_{4}R-i_{4}R+i_{5}R=0$ (3)

Recordemos tambien que:

$i_{1}=i_{2}+i_{3}$ (4)

$i_{2}=i_{4}+i_{5}$ (5)

Lo que debemos hacer ahora es resolver el sistema de ecuaciones y encontrar los valore de las corrientes. Por lo tanto, los valores de las corrientes serán:

$i_{1}=3/13\: A$