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3 years ago

According to Newton's second law of motion what is force equal to

2 answers:

8 0

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the netforce, and inversely proportional to the mass of the object

IrinaVladis [17]3 years ago

6 0

Explanation:

According to Newton's second law of motion, the force is equal to the product of mass and acceleration of the object. Mathematically, it is given by :

$F=m\times a$

It is also defined as the rate of change of linear momentum is directly proportional to the applied external force i.e.

$F=\dfrac{dp}{dt}$

If the momentum of the system is constant, the net force acting on it is equal to 0. Hence, this is the required solution.

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Thermodynamics

Akimi4 [234]

Answer:

E = 3.8 kJ

Explanation:

Given that,

The mass of the object, m = 10 g = 0.01 kg

The heat of fusion of aluminum is 380 kJ/kg

We need to find the energy required to melt the mass of the aluminium. It can be calculated as follows:

E = mL

So,

E = 0.01 × 380

E = 3.8 kJ

So, the energy required to melt the mass is equal 3.8 kJ.

7 0

3 years ago

Convert the following to relative uncertainties a) 2.70 ± 0.05cm b) 12.02 ± 0.08cm

DENIUS [597]

data which is expressed in form of following way

$a = a_o + \Delta a$

here in above expression

$a_o$ = true value

$\Delta a$ = uncertainty in the value

now the relative uncertainty is given as

$\frac{\Delta a}{a_o}$

now by above formula we can say

a) 2.70 ± 0.05cm

here

True value = 2.70

uncertainty = 0.05

Relative uncertainty = $\frac{0.05}{2.70}$ = 0.0185

b) 12.02 ± 0.08cm

here

True value = 12.02

uncertainty = 0.08

Relative uncertainty = $\frac{0.08}{12.02}$ = 0.00665

4 0

3 years ago

A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed

Tems11 [23]

Answer:

The box will be moving at 0.45m/s. The solution to this problem requires the knowledge and application of newtons second law of motion and the knowledge of linear motion. The vertical component of the force Fp acts vertically upwards against the directio of motion. This causes a constant upward force of 23sin45° to act on the box. Fhe frictional force of 13N also acts vertically upwards and so two forces act upwards against rhe force of gravity resulting un a net force of 0.7N acting kn the box. This corresponds to an acceleration of 0.225m/s². So in w.0s after i start to push v = 0.45m/s.

Explanation:

5 0

4 years ago

Read 2 more answers

A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The bo

WITCHER [35]

Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

$F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}$

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

$acceleration=\frac{30}{5}=6m/s^{2}$

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

$v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds$

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

5 0

4 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

jenyasd209 [6]

The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.

2. This is immediate from (2.2.7).

3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =

3 0

3 years ago