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2 years ago

8

a carpenter hits a nail with a hammer. compared to the magnitude of the force the hammer exerts on the nail, the magnitude of th

e force the nail exerts on the hammer during contact is?

2 answers:

Varvara68 [4.7K]2 years ago

8 0

Answer:

Equal (and the direction is opposite)

Explanation:

Newton's third law of motion states that:

"when an object A exerts a force on an object B, object B exerts an equal and opposite force on object A. The two forces are called action and reaction"

In this situation, we have a hammer exerting a force on a nail. We can think the hammer as object A and the nail as object B, so according to Newton's third law, the nail must exert an equal and opposite force on the hammer.

Therefore, the magnitude of the force that the nail exerts on the hammer is equal to the force exerted by the hammer on the nail (and in opposite direction)

mario62 [17]2 years ago

7 0

The exact same amount of force will be exerted but in opposite direction

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A block of mass m is pushed up against a spring with spring constant k until the spring has been compressed a distance x from eq

Snowcat [4.5K]

Answer:d

Explanation:

Spring is compressed to a distance of x from its equilibrium position

Work done by block on the spring is equal to change in elastic potential energy

i.e. Work done by block $W=\frac{1}{2}kx^2$

therefore spring will also done an equal opposite amount of work on the block in the absence of external force

Thus work done by spring on the block $W=-\frac{1}{2}kx^2$

Thus option d is correct

6 0

2 years ago

The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo

irina1246 [14]

Answer:

$p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

Explanation:

first write the newtons second law:

F $_{s}$ =δma $_{s}$

Applying bernoulli,s equation as follows:

∑ $δp+\frac{1}{2} ρδV^{2} +δγz=0\\$

Where, $δp$ is the pressure change across the streamline and $V$ is the fluid particle velocity

substitute $ρg$ for {tex]γ[/tex] and $g_{0}-cz$ for $g$

$dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0$

integrating the above equation using limits 1 and 2.

$\int\limits^2_1 \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

there the bernoulli equation for this flow is $p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

note: $ρ$ =density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

4 0

2 years ago

Lilli suggests that they explore the simulation starting with varying only a single parameter in order to understand the role of

mrs_skeptik [129]

Answer:

B.

Explanation:

One of the ways to address this issue is through the options given by the statement. The concepts related to the continuity equation and the Bernoulli equation.

Through these two equations it is possible to observe the behavior of the fluid, specifically the velocity at a constant height.

By definition the equation of continuity is,

$A_1V_1=A_2V_2$

In the problem $A_2$ is $2A_1$ , then

$A_1V_1=2A_1V_2$

$V_2 = \frac{V_1}{2}$

<em>Here we can conclude that by means of the continuity when increasing the Area, a decrease will be obtained - in the diminished times in the area - in the speed.</em>

For the particular case of Bernoulli we have to

$P_1 + \frac{1}{2}\rho V_1^2 = P_2 +\frac{1}{2}\rho V_2^2$

$P_2-P_1 = \frac{1}{2} \rho (V_1^2-V_2^2)$

For the previous definition we can now replace,

$P_2-P_1 = \frac{1}{2} \rho (V_1^2-(\frac{V_1}{2})^2)$

$\Delta P = \frac{3}{8} \rho V_1^2$

<em>Expressed from Bernoulli's equation we can identify that the greater the change that exists in pressure, fluid velocity will tend to decrease</em>

The correct answer is B: "If we increase A2 then by the continuity equation the speed of the fluid should decrease. Bernoulli's equation then shows that if the velocity of the fluid decreases (at constant height conditions) then the pressure of the fluid should increase"

4 0

2 years ago

Which law of motion accounts for the following statement? "When a marble and a billiard ball are impacted by the same force, the

Komok [63]

The second law explains this.

4 0

2 years ago

Read 2 more answers

A person is working on a steel structure while standing on the ground. An accident occurred where 5 A pass through the structure

matrenka [14]

Answer:

35mA

Explanation:

Hello!

To solve this problem we must use the following steps

1. Find the electrical resistance of the metal rod using the following equation

$R=\alpha \frac{l}{a}$

WHERE

α=

metal rod resistivity=2x10^-4 Ωm

l=leght=2m

A= Cross-sectional area

$A=\frac{\pi }{4} d^2=\frac{\pi }{4} (0.06)^2=0.00283$

solving

$R=(2x10^-4)\frac{2}{0.00283} =0.14$

2. Now we model the system as a circuit with parallel resistors, where we will call 1 the metal rod and 2 the man(see attached image)

3.we know that the sum of the currents in 1 and 2 must be equal to 5A, by the law of conservation of energy

I1+I2=5

4.as the voltage on both nodes is the same we can use ohm's law in resitance 1 and 2 (V=IR)

V1=V2

(0.14I1)=2000(i2)

solving for i1

I1=14285.7i2

5.Now we use the equation found in step 3

14285.7i2+i2=5

$i2=\frac{5}{14285.7+1} =3.5x10^-4A=35mA$

6 0

2 years ago