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Serjik [45]
3 years ago
9

a 2.6 kg block is attached to a horizontal spring that has a spring constant of 126 N/m. At the instant when the displacement of

the spring from its unstrained length is -0.115 m, what is the acceleration a of the object
Physics
1 answer:
NeTakaya3 years ago
8 0

Answer:

The acceleration of the object is 5.57\ m/s^2.

Explanation:

Given that,

Mass of the block, m = 2.6 kg

Spring constant of the spring, k = 126 N/m

At the instant when the displacement of the spring from its unstained length is 0.115 m. We need to find the acceleration of the object.

When the block is displaced, the force acting on the spring is equal to the force due to its motion. Such as :

kx=ma

a is acceleration of the object

a=\dfrac{kx}{m}\\\\a=\dfrac{126\times 0.115}{2.6}\\\\a=5.57\ m/s^2

So, the acceleration of the object is 5.57\ m/s^2.

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leva [86]

The force exerted on Alice’s foot as she kicks the ball is 1.25 N.Force can be defined as the influence that can accelerate the object.

<h3>What is force?</h3>

Force can be defined as the influence that can accelerate the object. From Newton's Second Law of motion,

F = ma

Where,

F - force

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a - acceleration = 5 N

Put the values in the formula,

F = 0.25 \times 5\\\\F = 1.25 \rm \ N

Therefore, the force exerted on Alice’s foot as she kicks the ball is 1.25 N.

Learn more about Force:

brainly.com/question/2840101

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When grocery shopping, the mass of the cart changes as you start to fill up your cart. How does the change in mass of your cart
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3 years ago
Two castings are proposed for a study. One casting is a cube 10.16 cm (4 in.) on a side; the other is a rectangle 10.64 cm (16 i
iren [92.7K]

To solve the problem it is necessary to apply the concepts related to Chvorinov's Law, which states that

\frac{T_r}{T_c} = (\frac{V_r}{V_c}*\frac{SA_c}{SA_r})^2

Where,

V_c = Volume cube

SA_c = Superficial Area from Cube

V_r = Volume Rectangle

SA_r= Superficial Area from Rectangle

Our values are given as (I will try to develop the problem in English units for ease of calculations),

V_c = 4^3 = 64in^3

SA_c = 6*4^2=96in^2

SA_r = 2*(1*16+4*1+16*4)=168in^2

V_r = 4*1*16 = 64in^3

Applying the Chvorinov equation we have to,

\frac{T_r}{T_c} = (\frac{V_r}{V_c}*\frac{SA_c}{SA_r})^2

\frac{T_r}{T_c} = (\frac{64}{64}*\frac{96}{168})^2

\frac{T_r}{T_c} = (\frac{96}{168})^2

\frac{T_r}{T_c} = 0.3265

The stipulated time for the cube is 14.5 then,

T_r = 0.3265*14.5

T_r = 4.735min

4 0
4 years ago
The next two questions refer to this situation: A rectangular loop with sides of length a= 1.00 cm and b= 2.70 cm is placed near
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Answer:

a) 1.007 * 10^{-7} V.m , into the page

b) -5.39 * 10^{-8} V, counterclockwise

Explanation:

a) d\Phi = B.dA\\\\B_r = \frac{\mu_0I}{2\pi r}\\\\dA = b.dr\\\\\Phi = \int\limits^b_a {B.dA} = \int\limits^{a+d}_d {\frac{\mu_0I}{2\pi r}.b.dr} =\frac{\mu_0I}{2\pi }.b.ln(\frac{a+d}{d} )\\

At t = 2.90, I = 3.62 + 1.49 * (2.90)^2 = 16.15 A

\Phi = \frac{4\pi * 10^{-7}*16.15*0.027}{2\pi} ln(1.46/0.46) = 1.007 * 10^{-7} V.m

Direction is into the page.

b) emf = -d\Phi/dt = -\frac{\mu_0}{2\pi }.b.ln(\frac{a+d}{d}).(2.98t)= \\=-\frac{4\pi * 10^{-7}*0.027}{2\pi} ln(1.46/0.46)*2.98*2.90= -5.39 * 10^{-8} V

Direction is counterclockwise.

8 0
3 years ago
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