List all the formulae in Physics

the (speed) of an object is a vector quantity that tells us the speed and direction of an objects momentum is this correct? If n

Ivenika [448]

Answer:

sorry kailangan points wh

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2 years ago

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An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

$v_{4} = -6\,\frac{m}{s}$

Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

3 0

3 years ago

The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca

d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

$U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}$

the elastic potencial energy is given by:

$U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J$

The work is defined as:

$W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J$

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

$U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J$

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

$8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s$

3 0

3 years ago

Why would knowing the characteristics of circuits be important in designing electrical circuits?

zubka84 [21]

Well if you didn't you could make mistakes, which would lead ,in the best case, at a fail of the circuit , or if it goes out of control it could be dangerous

for example you have to know that the wires become hot and they loose their abbilitys as connecters(the hotter it will, the more energy you lose becouse the R will be bigger)

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3 years ago

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What keeps both the cars pressed down on the road?

mihalych1998 [28]

Answer:

Gravity

Explanation:

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3 years ago