A geologist is trying to determine what time period a layer of rock was formed.

A polar bear runs at a speed of 11 m/s and has a mass of 380.2 kg. How much Kinetic energy does the bear have?

Yanka [14]

Answer:

$\boxed{\sf Kinetic \ energy \ of \ the \ bear (KE) = 23002.1 \ J}$

Given:

Mass of the polar bear (m) = 6.8 kg

Speed of the polar bear (v) = 5.0 m/s

To Find:

Kinetic energy of the polar bear (KE)

Explanation:

Formula:

$\boxed{ \bold{\sf KE = \frac{1}{2} m {v}^{2} }}$

Substituting values of m & v in the equation:

$\sf \implies KE = \frac{1}{2} \times 380.2 \times {11}^{2}$

$\sf \implies KE = \frac{1}{ \cancel{2}} \times \cancel{2} \times 190.1 \times 121$

$\sf \implies KE = 190.1 \times 121$

$\sf \implies KE = 23002.1 \: J$

$\therefore$

Kinetic energy of the polar bear (KE) = 23002.1 J

5 0

3 years ago

Which one of the following indicates a drop in temperature?

Nitella [24]

Answer: <span>D. A bimetallic strip bends so that the steel is on the outside curve
</span>
When something has an increased temperature, its volume will expand. Then, if the temperature drops, its volume should be smaller. From there option A and B are out since the liquid in thermometer is expand or move up.

When you put two kinds of different metal with a different coefficient of thermal expansion, the outer curve metal will be the one with lesser coefficient when temperature drop. Since the question about drop in temperature then the metal should be bend
Brass will expand 1.5 times more than the steel so the outer curve would be the steel.

6 0

4 years ago

which statement best describes how changes behave a.opposites charges repel against each other and like charges attract to each

Anon25 [30]

I think it is B hope this helps

God Bless

7 0

3 years ago

Read 2 more answers

Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B i

mel-nik [20]

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

$\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2} =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}$

$\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad$

$w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}$

The velocity at the contact point is equal to:

$v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}$

$v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}$

Matching both expressions:

$1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s$

b) The time during which the disks slip is:

$t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s$

a) The angular acceleration of each disk is

$\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)$

$\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)$

6 0

3 years ago

If you pull horizontally on a desk with a force of 150 N and the desk doesn't move, the friction force must be 150 N. Now if you

s344n2d4d5 [400]

Answer:

The friction force is 250 N

Explanation:

The desk is moving at constant velocity. This means that its acceleration is zero: a = 0. Newton's second law states that the resultant of the forces acting on the desk is equal to the product between mass (m) and acceleration (a):

$\sum F=ma$

In this case, we know that the acceleration is zero: a = 0, so also the resultant of the forces must be zero:

$\sum F = 0$ (1)

We are only interested in the forces acting along the horizontal direction, since it is the direction of motion. There are two forces acting in this direction:

- the pull, forward, F = 250 N

- the friction force, backward, $F_f$

Given (1), we have

$F-F_f = 0$

So the force of friction must be equal to the pull:

$F=F_f = 250 N$

8 0

3 years ago