In which of the following is positive work done by a person on a suitcase

A rock that has magnetic properties is

Dahasolnce [82]

Answer: lodestone

Explanation:

ITS LODESTONE

6 0

2 years ago

Can someone help<br> pls !

Gemiola [76]

A sort of electricity is a light bulb or a phone / computer charger. plants food water. the sun and rain . that’s what i’m guessing!

6 0

3 years ago

A uniform sphere (I = 2/5 MR 2 ) rolls down an incline. (a) What must be the incline angle if the linear acceleration of the cen

Liono4ka [1.6K]

Answer:

Part a)

$\theta = 8.05 degree$

Part b)

$a = 1.37 m/s^2$

Explanation:

As the uniform sphere is rolling down the inclined plane then the net force on the sphere is given as

$mg sin\theta - F_f = ma$

also we have torque equation on it

$F_f R = I\alpha$

for pure rolling

$a = R \alpha$

$F_f = \frac{Ia}{R^2}$

now we have

$mg sin\theta = ma + \frac{Ia}{R^2}$

now we have

$mg sin\theta = (m + \frac{2}{5}m)a$

$a = \frac{5}{7}g sin\theta$

now given that

$a = 0.10 g$

so we have

$0.10 g = \frac{5}{7} g sin\theta$

$sin\theta = 0.14$

$\theta = 8.05 degree$

Part b)

If the inclined plane is frictionless then the acceleration is given as

$a = g sin\theta$

$a = 9.8(0.14)$

$a = 1.37 m/s^2$

5 0

3 years ago

Read 2 more answers

Researchers in the Antarctic measure the temperature to be -32°F. What is this temperature on the following scales?(a) the Celsi

Illusion [34]

Answer:

a) -35.6°C

b) 237.4 K

Explanation:

To convert temperature from degree celsius to degree fahrenheit, use the formula below:

$T_c=\frac{5}{9}(T_f-32)$

a) Therefore to convert -32°F to celsius, substitute it into the celsius

$\begin{gathered} T_c=\frac{5}{9}(-32-32) \\ \\ T_c=\frac{5}{9}(-64) \\ \\ T_c=-35.6^0C \end{gathered}$

b) To covert to the Kelvin scale, use the formula below

$\begin{gathered} T_k=T_c+273 \\ \\ T_k=-35.6+273 \\ \\ T_k=237.4K \\ \end{gathered}$

8 0

1 year ago

A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 110,000 kg of scrap metal in

erica [24]

Explanation:

Given that,

Mass of a freight car, $m_1=30,000-kg$

Speed of a freight car, $u_1=0.85\ m/s$

Mass of a scrap metal, $m_2=110,000\ kg$

(a) Let us assume that the final velocity of the loaded freight car is V. The momentum of the system will remain conserved as follows :

$30000\times 0.85=(30000+110000)V\\\\V=\dfrac{30000\times 0.85}{30000+110000}\\\\=0.182\ m/s$

So, the final velocity of the loaded freight car is 0.182 m/s.

(b) Lost on kinetic energy = final kinetic energy - initial kinetic energy

$\Delta K=\dfrac{1}{2}[(m_1+m_2)V^2-m_1u_1^2)]\\\\=\dfrac{1}{2}\times [(30,000+110,000 )0.182^2-30000(0.85)^2]\\\\=-8518.82\ J$

Lost in kinetic energy is 8518.82. Negative sign shows loss.

6 0

2 years ago