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3 years ago

How does mechanical energy relate to the work an object can do

Physics

1 answer:

kifflom [539]3 years ago

6 0

I am 11 year old and u just needed to login in so I can't answer ur questions

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A small wooden block with mass 0.750 kg is suspended from the lower end of a light cord that is 1.72 m long. The block is initia

Ierofanga [76]

Answer:

$v_{0}=319.2 m/s$

Explanation:

We need to use the momentum and energy conservation.

$p_{0}}=p_{f}$

$mv_{0}=(m+M)V_{1}$

Where:

m is the mass of bullet (m=0.01 kg)
M is the mass of the wooden (M=0.75 kg)
v(0) initial velocity of bullet
V(1) is the velocity of the combined object in the moment the bullet hist the block

Conservation of energy.

We have kinetic energy at first and kinetic and potential energy at the end.

$(1/2)(m+M)V_{1}^{2}=(1/2)(m+M)V_{2}^{2}+(m+M)gh$

Here:

V(1) is the velocity of the combined at the initial position
h is the vertical height ( h = 0.800 m)

We can find V(2) using the definition of force at this point:

$\Sigma F=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)$

$T-(m+M)gcos(\theta)=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)$

$cos(\theta) =(L-h)/L=(1.72-0.8)/1.72$

$\theta =57.66$

Now, we can solve the equation to find V(2)

$V_{2}=\sqrt{\frac{R*(T-(m+M)*g*cos(\theta))}{(m+M)}}$

$V_{2}=\sqrt{\frac{1.72*(4.86-(0.01+0.75)*9.81*cos(57.66))}{(0.01+0.75)}}$

$V_{2}=1.40 m/s$

Now we can find V(1) using the conservation of energy equation

$(1/2)V_{1}^{2}=(1/2)V_{2}^{2}+gh$

$V_{1}=\sqrt{V_{2}^{2}+2gh}$

$V_{1}=\sqrt{1.40^{2}+2*9.81*0.8}$

$V_{1}=4.20 m/s$

Finally, using the momentum equation we find v(0)

$v_{0}=\frac{(m+M)V_{1}}{m}$

$v_{0}=\frac{(0.01+0.75)*4.20}{0.01}$

$v_{0}=319.2 m/s$

I hope it helps you!

7 0

4 years ago

A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.

mafiozo [28]

Answer:

a. $f=1.22*10^{-15} N$

b. $f=53.6*10^{-17} N$

Explanation:

The force existing between two charges is given as

$f=\frac{kq_{1}q_{2}}{r^{2}}$

where q= charge,

k=constant

r= distance between the two charges

Note: this force can either be repulsive or attractive force depending on the charges involve. it is repulsive if they are similar charge and it is attractive if it is opposite charges.

Also the charge of an electron is

$-1.602*10^{-19}$

A. we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{0.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{0.04}\\f_{21}=1.44*10^{-15}Ni$

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis

for the -5.50nC the distance between them is 0.600m as can be seen in the diagram the magnitude of the force is

$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.6^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.36}\\f_{23}=-(0.22*10^{-15})N i$

this this force will be repulsive force and it points away from the electron i.e points towards the -ve x-axis.

The total net force on the electron is thus

$f=f_{21}+f_{23}\\ f=1.44*10^{-15}-0.22*10^{-15}\\ f=1.22*10^{-15} N$

b. at distance of x=1.20m, this is shown on the diagram below (attachment 2)

we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{1.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{1.44}\\f_{21}=4.0*10^{-17}Ni$

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

for the -5.50nC the distance between them is 0.4m as can be seen in the diagram the magnitude of the force is

$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.4^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.16}\\f_{23}=49.6*10^{-17}Ni$

this this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

The total net force on the electron is thus

$f=f_{21}+f_{23}\\ f=4.0*10^{-15}+49.6*10^{-17}\\ f=53.6*10^{-17} N$

8 0

3 years ago

Which statements about electric field lines are correct? Check all that apply.

slava [35]

Answer:

they cross over one another between charge.

7 0

4 years ago

A soccer goal is 2.44 m high. A player kicks the ball at a distance 13 m from the goal at an angle of 40°, and the ball just hit

Vera_Pavlovna [14]

Let v denote the initial speed of the ball. The ball's position at time t is given by the vector

$\mathbf r(t)=v\cos40^\circ\,t\,\mathbf i+\left(v\sin40^\circ\,t-\dfrac g2t^2\right)\,\mathbf j$

where g is the acceleration due to gravity with magnitude 9.80 m/s^2.

The ball reaches the goal 13 m away at time t such that

$10\,\mathrm m=v\cos40^\circ t\implies t=\dfrac{10\,\mathrm m}{v\cos40^\circ}$

at which point it attains a height of 2.44 m, so that

$2.44\,\mathrm m=v\sin40^\circ\left(\dfrac{10\,\mathrm m}{v\cos40^\circ}\right)-\dfrac g2\left(\dfrac{10\,\mathrm m}{v\cos40^\circ}\right)^2$

$2.44\,\mathrm m=(10\,\mathrm m)\tan40^\circ-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)\left(\dfrac{100\,\mathrm m^2}{v^2\cos^240^\circ}\right)$

$\implies\boxed{v\approx3.75\dfrac{\rm m}{\rm s}}$

6 0

4 years ago

Tom has a mass of 73 . 9 kg and Sally has a mass of 59 . 8 kg. Tom and Sally are standing 27 . 9 m apart on a massless dance flo

Solnce55 [7]

Gravitational force(F) = Gxm1xm2 / r ^2. F = 6 . 67259 × 10 ^− 11 x 73.9 x 53.8 / (27.9)^2 = 3.408x10^-10N which is negligible. Hope it is clear.

3 0

3 years ago