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3 years ago

When a car stops, the brakes heat up because of friction. What is this an example of?

Physics

1 answer:

Umnica [9.8K]3 years ago

6 0

Kinetic Energy would have to be the answer. Pretty sure.

You might be interested in

A tightrope walker more easily balances on a tightwire if his pole

cestrela7 [59]

B) droops.

Why?
To maintain balance, you do not need something short so you're balanced well... You need something long and droopy to maintain balance. The pole should be held by your waist and it should be light.

Hope this helps!~

4 0

2 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0

2 years ago

The volume of liquid flowing per second is called the volume flow rate Q and has the dimensions of [L]3/[T]. The flow rate of li

melamori03 [73]

Answer: n=4

Explanation:

We have the following expression for the volume flow rate $Q$ of a hypodermic needle:

$Q=\frac{\pi R^{n}(P_{2}-P_{1})}{8\eta L}$ (1)

Where the dimensions of each one is:

Volume flow rate $Q=\frac{L^{3}}{T}$

Radius of the needle $R=L$

Length of the needle $L=L$

Pressures at opposite ends of the needle $P_{2}$ and $P_{1}=\frac{M}{LT^{2}}$

Viscosity of the liquid $\eta=\frac{M}{LT}$

We need to find the value of $n$ whicha has no dimensions, and in order to do this, we have to rewritte (1) with its dimensions:

$\frac{L^{3}}{T}=\frac{\pi L^{n}(\frac{M}{LT^{2}})}{8(\frac{M}{LT}) L}$ (2)

We need the right side of the equation to be equal to the left side of the equation (in dimensions):

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n}}{LT}$ (3)

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n-1}}{T}$ (4)

As we can see $n$ must be 4 if we want the exponent to be 3:

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{4-1}}{T}$ (5)

Finally:

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{3}}{T}$ (6)

8 0

3 years ago

Two boys are sliding toward each other on a frictionless, ice-covered parking lot. Jacob, mass 45 kg, is gliding to the right at

Maurinko [17]

Answer:

$v=0.3381\ m.s^{-1}$ in the direction of motion of Jacob

Explanation:

Given:

mass of Jacob, $m_j=45\ kg$

velocity of Jacob, $v_j=8.08\ m.s^{-1}$

mass of Ethan, $m_e=31\ kg$

velocity of Ethan, $v_e=10.9\ m.s^{-1}$

Now using the conservation of linear momentum for the case:

(When the two masses in motion combine to form one after the collision then they will move together in the direction of the greater momentum.)

$(m_j+m_e).v=m_j.v_j-m_e.v_e$

$(45+31)\times v=45\times 8.08-31\times 10.9$

$v=0.3381\ m.s^{-1}$ in the direction of motion of Jacob as it was assumed to be positive.

5 0

3 years ago

When a force of 450N pushes on a 20kg box as

ehidna [41]

Answer:

ma+mgsinh0+f=F∴(25)(0.75)+(25)(10)sinh0+μkN=F∴18.75+(250)(0.6h)+μk(mgcosh0=F⟹18.75+150+μk((25)(10)(0.76))=500∴168.75+μk(190)=500⟹μk(190)=331.25⟹μk=1.74

Explanation:

7 0

2 years ago