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2 years ago

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Block A of mass 2.0 kg is released from rest at the top of a 3.6 m long plane inclined at an angle of 30o, as shown in the figur

e above. After sliding on the horizontal surface, block A hits and sticks to block B, which is at rest and has mass 3.0 kg. Assume friction is negligible. The speed of the blocks after the collision is most nearly

2 answers:

Jet001 [13]2 years ago

4 0

Answer:

............................

31kg

Helen [10]2 years ago

4 0

Answer:

2.4 m/s

Explanation:

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A reversible Carnot engine has an efficiency of 30% and it operates between a heat source maintained at T 1 , and a refrigerator

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To solve this problem we will apply the concepts related to the thermal efficiency given in an engine of the Carnot cycle. Here we know that efficiency is given under the equation

$\eta = 1 - \frac{T_2}{T_1}$

Where,

$T_2 =$ Temperature of Cold Body

$T_1 =$ Temperature of Hot Body

$\eta$ = Efficiency

According to the statement our values are:

$\eta = 0.3$

$T_2 = 210K$

Replacing we have that

$0.3 = 1- \frac{210}{T_1}$

$\frac{210}{T_1} = 0.7$

$T_1 = \frac{210}{0.7}$

$T_1 = 300K$

Therefore the temperature of the heat source is 300K

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1 year ago

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Why are there only two elements in the first period of the periodic table?(1 point)

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2 years ago

please help! find magnitude and direction (the counterclockwise angle with the +x axis) of a vector that is equal to a + c

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Answer:

Option (2)

Explanation:

From the figure attached,

Horizontal component, $A_x=A\text{Sin}37$

$A_x=12[\text{Sin}(37)]$

= 7.22 m

Vertical component, $A_y=A[\text{Cos}(37)]$

= 9.58 m

Similarly, Horizontal component of vector C,

$C_x$ = C[Cos(60)]

= 6[Cos(60)]

= $\frac{6}{2}$

= 3 m

$C_y=6[\text{Sin}(60)]$

= 5.20 m

Resultant Horizontal component of the vectors A + C,

$R_x=7.22-3=4.22$ m

$R_y=9.58-5.20$ = 4.38 m

Now magnitude of the resultant will be,

From ΔOBC,

$R=\sqrt{(R_x)^{2}+(R_y)^2}$

= $\sqrt{(4.22)^2+(4.38)^2}$

= $\sqrt{17.81+19.18}$

= 6.1 m

Direction of the resultant will be towards vector A.

tan(∠COB) = $\frac{\text{CB}}{\text{OB}}$

= $\frac{R_y}{R_x}$

= $\frac{4.38}{4.22}$

m∠COB = $\text{tan}^{-1}(1.04)$

= 46°

Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.

Option (2) will be the answer.

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2 years ago