Answer:<br><br> A. 2y(x³ + 9x - 5x² - 45)<br><br> B. 2y(x² + 9)(x

Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. It has a centripetal force. F acting on it. If bod

cluponka [151]

Answer:

The centripetal force on body 2 is 8 times of the centripetal force in body 1.

Explanation:

Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. The centripetal force acting on it is given by :

$F=\dfrac{mv^2}{r}$

Body 2 has a mass 2m and its moving in a circle of radius 4r at a speed 4v. The centripetal force on body 2 is :

$F'=\dfrac{2m\times (4v)^2}{4r}\\\\F'=\dfrac{2m\times 16v^2}{4r}\\\\F'=8\dfrac{mv^2}{r}\\\\F'=8F$

So, the centripetal force on body 2 is 8 times of the centripetal force in body 1.

8 0

3 years ago

Which type of macromolecules a cell brake down food?

kobusy [5.1K]

Answer:

Hope this helps you find the answer

Explanation:

The proteins, lipids, and polysaccharides that make up most of the food we eat must be broken down into smaller molecules before our cells can use them—either as a source of energy or as building blocks for other molecules.

8 0

3 years ago

The minute hand of a wall clock measures 16 cm from its tip to the axis about which it rotates. The magnitude and angle of the d

olya-2409 [2.1K]

Answer:

Explanation:

Given

Minute hand length =16 cm

Time at a quarter after the hour to half past i.e. 1 hr 45 min

Angle covered by minute hand in 1 hr is 360 and in 45 minutes 270

$|r|=\frac{3\times 2\pi r}{4}=75.408 cm$

$Angle =270^{\circ}$

(c)For the next half hour

Effectively it has covered 2 revolution and a quarter

$|r|=\frac{2\pi r}{4}=25.136 cm$

angle turned $=90^{\circ}$

(f)Hour after that

After an hour it again comes back to its original position thus displacement is same =25.136

Angle turned will also be same i.e. $90 ^{\circ}$

7 0

4 years ago

The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s.

sattari [20]

Answer:

$d=691.71km$

Explanation:

The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:

$t=\frac{d}{v_t}-\frac{d}{v_l}$

Here d is the distance at which the earthquake take place and $v_t, v_l$ is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:

$t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km$

8 0

3 years ago

Time left 0:43:35

ss7ja [257]

Answer:

Bro where is image Hope you understand me

7 0

3 years ago

Answer: A. 2y(x³ + 9x - 5x² - 45) B. 2y(x² + 9)(x - 5)