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almond37 [142]
3 years ago
7

A small block of mass m on a horizontal frictionless surface is attached to a horizontal spring that has force constant k. The b

lock is pushed against the spring, compressing the spring a distance d. The block is released, and it moves back and forth on the end of the spring. During its motion, what is the maximum speed of the block?
Physics
1 answer:
Rashid [163]3 years ago
5 0

Answer:

v_{max} = |d|\cdot \sqrt{\frac{k}{m} }

Explanation:

The maximum speed of the block occurs when spring has no deformation, that is, there is no elastic potential energy, which can be remarked from appropriate application of the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot d^{2} = \frac{1}{2}\cdot m \cdot v^{2}

k \cdot d^{2} = m\cdot v^{2}

v_{max} = |d|\cdot \sqrt{\frac{k}{m} }

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According to Ohm’s law, which combination of units is the same as the unit for resistance? volt ÷ ampere ampere × volt volt + am
Maksim231197 [3]

Answer:

volt ÷ ampere

Explanation:

The mathematical form of Ohms law is given by :

V = IR

Where V is voltage

I is current

R is resistance

R=\dfrac{V}{I}

The unit of voltage is volt and that of current is ampere

Unit of resistance :

R=\dfrac{\text{volt}}{\text{ampere}}

So, volt ÷ ampere is the same as the unit of resistance. Hence, the correct option is (a).

6 0
2 years ago
Read 2 more answers
CAN someone help ASAP?
lyudmila [28]

Answer:someone help me

Explanation:

7 0
3 years ago
A 15 kg mass is moving at 7.50 meters per second on a horizontal, frictionless surface. What is the total work that must be done
sashaice [31]
Kinetic energy = (1/2) (mass) x (speed)²

At 7.5 m/s, the object's KE is (1/2) (7.5) (7.5)² = 210.9375 joules

At 11.5 m/s, the object's KE is (1/2) (7.5) (11.5)² = 495.9375 joules

The additional energy needed to speed the object up from 7.5 m/s
to 11.5 m/s is (495.9375 - 210.9375) = <em>285 joules</em>.

That energy has to come from somewhere. Without friction, that's exactly
the amount of work that must be done to the object in order to raise its
speed by that much.
8 0
3 years ago
A person doing chin-up weighs 700.0 N, disregarding the weight of the arms. During the first 25.0 cm of the lift, each arm exert
dlinn [17]
First, we will get the resultant force:
The direction of the force due to the person's weight is vertically down.
weight of person = 700 newton

Assume that the force exerted by the arms has a vertically upwards direction.
Force exerted by arms = 2*355 = 710 newtons

Therefore, the resultant force = 710 - 700 = 10 newtons (in the vertically upwards direction) 

Now, we will get the mass of the person.
weight = 700 newtons
weight = mass * acceleration due to gravity
700 = 9.8*mass
mass = 71.428 kg

Then we will calculate the acceleration of the resultant force:
Force = mass*acceleration
10 = 71.428*acceleration
acceleration = 0.14 m/sec^2

Finally, we will use the equation of motion to get the final speed of the person.
V^2 = U^2 + 2aS where:
V is the final velocity that we need to calculate
U is the initial velocity = 0 m/sec (person starts at rest)
a is the person's acceleration = 0.14 m/sec^2
S is the distance covered = 25 cm = 0.25 meters

Substitute with the givens in the above equation to get the final speed as follows:
V^2 = U^2 + 2aS
V^2 = (0)^2 + 2(0.14)(0.25)
V^2 = 0.07
V = 0.2645 m/sec

Based on the above calculations:
The person's speed at the given point is 0.2645 m/sec

4 0
3 years ago
Read 2 more answers
An ideal spring hangs from the ceiling. A 1.95 kg mass is hung from the spring, stretching the spring a distance d=0.0865 m from
uranmaximum [27]

Answer:

kinetic energy = 0.1168 J

Explanation:

From Hooke's law, we know that ;

F = kx

k = F/x

We are given ;

Mass; m = 1.95 kg

Spring stretch; d = x = 0.0865

So, Force = mg = 1.95 × 9.81

k = 1.95 × 9.81/0.0865 = 221.15 N/m

Now, initial energy is;

E1 = mgL + ½k(x - L)²

Also, final energy; E2 = ½kx² + ½mv²

From conservation of energy, E1 = E2

Thus;

mgL + ½k(x - L)² = ½kx² + ½mv²

Making the kinetic energy ½mv² the subject, we have;

½mv² = mgL + ½k(x - L)² - ½kx²

We are given L=0.0325 m

Plugging other relevant values, we have ;

½mv² = (1.95 × 9.81 × 0.0325) + (½ × 221.15(0.0865 - 0.0325)² - ½(221.15 × 0.0865²)

½mv² = 0.62170875 + 0.3224367 - 0.82734979375

½mv² = 0.1168 J

7 0
3 years ago
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