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Sever21 [200]
3 years ago
6

Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proport

ional to its mass m. That is, a = F / mWhat are the dimensions of force?
Physics
1 answer:
netineya [11]3 years ago
5 0

Answer:

[F]=[MLT^{-2}]

Explanation:

Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proportional to its mass m. The mathematical expression for the second law of motion is given by :

F = m × a

F is the applied force

m is the mass of the object

a is the acceleration due to gravity

We need to find the dimensions of force. The dimension of force m and a are as follows :

[m]=[M]

[a]=[LT^{-2}]

So, the dimension of force F is, [F]=[MLT^{-2}]. Hence, this is the required solution.

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A race car travels 40 m/s around a banked (45° with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of
OLEGan [10]

Answer:

c)F_{net} = 0.640 kN

Explanation:

As we know that resultant force is the net force that is acting on the system

As per Newton's II law we know that net force is product of mass and acceleration

so we will have

F_{net} = ma

here we know

m = 80 kg

for circular motion acceleration is given as

a_c = \frac{v^2}{R}

a_c = \frac{40^2}{200} = 8 m/s^2

now we have

F_{net} = 80 \times 8

F_{net} = 640 N

F_{net} = 0.640 kN

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3 years ago
¿cual es la velocidad de un auto que recorre 4566 metros en 4 minutos? expresar en km\h
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Explanation:

(4566 m / 4 min) × (1 km / 1000 m) × (60 min / h) = 68.49 km/h

3 0
3 years ago
A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M a
aleksklad [387]

Answer:

a)  y = 0.98 t², t=1s y= 0.98 m,  

b) he two blocks must move the same distance

c) v = 1.96 m / s,  d)  a = -1.96 m / s², e)  x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

             N-W = 0

             N = M g

X axis

             T- fr = Ma

the friction force has the expression

             fr = μ N

             fr = μ Mg

small block

             w- T = m a

             

we write the system of equations

             T - fr = M a

             mg - T = m a

we add and resolved

             mg-  μ Mg = (M + m) a

             a = g \ \frac{m - \mu M}{m+M}

             a = 9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}

             a = 9.8 (6/30)

             a = 1.96 m / s²

a) now we can use the kinematic relations

             y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

             y = ½ a t²

             y = ½ 1.96 t²

             y = 0.98 t²

for t = 1s y = 0.98 m

       t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

           v = vo + a t

           v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

       t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

           -fr = M a

          - μ M g = M a

          a = - μ g

          a = - 0.2 9.8

          a = -1.96 m / s²

e) the distance to stop the block is

         v² = vo² - 2 a x

        0 = vo² - 2a x

        x = vo² / 2a

        x = 1.96² / 2 1.96

        x = 0.98 m

the time to travel this distance is

        v = vo - a t

        t = vo / a

        t = 1.96 /1.96

        t = 1 s

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3 years ago
What is the lens equation?
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7 0
3 years ago
Read 2 more answers
Each plate of a parallel‑plate capacitor is a square of side 3.63 cm, and the plates are separated by 0.473 mm. The capacitor is
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Answer:

E= 55.53 x 10³ V/m

Explanation:

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a=  3.63 cm

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distance ,d= 0.473 mm

Stored energy ,U = 8.49 nJ

Value of capacitor given as

C=\dfrac{\varepsilon _oA}{d}

By putting the values

C=\dfrac{8.85\times 10^{-12}\times 3.63^2\times 10^{-4}}{0.473\times 10^{-3}}

C=2.46 x 10⁻¹¹ F

U=\dfrac{1}{2}CV^2

V=Voltage difference

V=\sqrt{\dfrac{2U}{C}}

V=\sqrt{\dfrac{2\times 8.49\times 10^{-9}}{2.46\times 10^{-11}}}

V=26.27 V

V= E d

E=Electric filed

26.27 = E x 0.473 x 10⁻³

E= 55.53 x 10³ V/m

7 0
3 years ago
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