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Calculate (A⃗ ×B⃗ )⋅C⃗ for the three vectors A⃗ with magnitude A = 4.86 and angle θA = 23.5 ∘ measured in the sense from the +x

trasher [3.6K]

Answer:

(A⃗ ×B⃗ )⋅C⃗ = 69.868

Explanation:

We simplify the cross product first, thereafter the solution of the cross product is now simplified with the dot product as shown in the step by step calculation in the attachment

4 0

3 years ago

39 liters of water at

Harman [31]

Answer:

Temperature of the mixture=8 °C

Explanation:

Let, Temperature of the mixture=t °C

Q₁ = Q₂

=> m₁SΔΘ₁ = m₂SΔΘ₂

=> 39 × (20+t) = 21 × (60-t)

∴ t = 8 °C

hope you have understood this

pls mark it as the brainliest

3 0

3 years ago

At which of the following points does a roller coaster have the most potential energy? As it is going down a hill. At the top of

vova2212 [387]

Answer:

At the top of the hill.

Explanation:

As the roller coaster goes up the hill, kinetic energy (K.E) decreases, gravitational potential energy (G.P.E) increases .

As it reach the top of the hill, K.E becomes zero and G.P.E reaches maximum .

As it goes down the hill, K.E starts to increase and G.P.E decrease .

At the bottom of the hill, K.E reaches maximum and G.P.E becomes zero .

(Correct me it I am wrong)

6 0

3 years ago

Tarzan swings on a 31.0 m long vine initially inclined at an angle of 36.0◦ with the vertical. The acceleration of gravity if 9.

Gennadij [26K]

Answer:

$v=10.777m/s$

Explanation:

Tarzan swing can be thought of as change in potential energy by going from higher location We solve for height of beginning of the swing by using simple cosine equation:

So

$31Cos36=25.08\\E_{potential}=mgh\\$

ΔE=mg(h₂-h₁)

$=m*9.81(31-25.08)\\$

The potential energy of Tarzan initial position is converted into kinetic energy of his swing.By using kinetic equation

$E_{kinectic}=P_{potential}\\1/2mv^{2}=m*9.81(31.0-25.08)\\(1/2)v^{2}=9.81(31.0-25.08)\\0.5v^{2}=58.07\\v^{2}=58.07/0.5\\v=\sqrt{58.07/0.5}\\ v=10.777m/s$

8 0

3 years ago

A light rope is attached to a block with mass 3.60 kg that rests on a frictionless, horizontal surface. The horizontal rope pass

Readme [11.4K]

Answer and Explanation:

(a) The fre-body diagrams for each block is shown below. In the block of mass 3.60 kg, there are 3 forces acting on it: horizontal force due to the rope ( $F_{t}$ ), vertical gravitational force ( $F_{g}$ ) and vertical normal force ( $F_{n}$ ), due to the surface. Since there is no vertical movement, $F_{g}$ and $F_{n}$ cancels it out. So, for this block, net force is horizontal due to the rope $F_{t}$ .

The block of mass m is hanging from the pulley, so there is the force of the rope ( $F_{t}$ ) and the gravitational force ( $F_{g}$ ). Both are vertical, because there is no surface "holding" block m.

(b) Since both blocks are attached to each other, the acceleration will be the same. To calculate it, we use the Second Law of Motion:

$F_{r}=m.a$

$a=\frac{F_{r}}{m}$

$a=\frac{18.8}{3.6}$

a = 5.22

The acceleration of either block is 5.22 m/s².

(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².

Suppose positive referential is going up. To determine mass:

$F_{r}=m.a$

$F_{t}-F_{g}=m.a$