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3 years ago

13

Use the definition of scalar product, a with arrow · b with arrow = ab cos θ, and the fact that a with arrow · b with arrow = ax

bx + ayby + azbz to calculate the angle between the two vectors given by a with arrow = 5.0i hat + 5.0j + 4.0k and b with arrow = 6.0i hat + 3.0j + 5.0k.

1 answer:

Zolol [24]3 years ago

3 0

Answer:

5+5=10

4+6=10

angle between is 45

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Muscular Endurance is the ability of a muscle to continue to perform without fatigue.

Art [367]

Answer:

True.

Explanation:

Defenintion of Muscular Endurance:

The ability of a muscle (or set of muscles) to perform a repeated action without tiring.

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3 years ago

What happens to the position of an object as an unbalanced force acts on?

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Answer:

Unbalanced forces change the motion of an object. If an object is at rest and an unbalanced force pushes or pulls the object, it will move. Unbalanced forces can also change the speed or direction of an object that is already in motion.

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3 years ago

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A railroad car of mass 2.50*10^4 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroa

NemiM [27]

Answer:

(a) 2.5 m/s

(b) 37.5 KJ

Explanation:

(a)

From the law of conservation of momentum, Initial momentum=Final momentum

$mV_1+3mV_2=(m+3m)V_f=4mV_f$

$V_1+3V_2=4V_f$ and making $V_f$ the subject then

$V_f=\frac {V_1+3V_2}{4}$ and since $V_1$ is initial velocity of car, value given as 4 m/s, $V_2$ is the initial velocity of the three cars stuck together, value given as 2 m/s and $v_f$ is the final velocity which is unknown. By substitution

$V_f=\frac {4+(3\times2)}{4}=2.5 m/s$

(b)

Initial kinetic energy is given by

$\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ$

Final kinetic energy is given by

$\frac {4mV_f^{2}}{2}=\frac {4\times 2.5\times 10^{4}\times 2.5^{2}}{2}=312.5\times 10^{3} J=312.5 KJ$

The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence

Energy lost=350-312.5=37.5 KJ

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3 years ago

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Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

26. A 40 kg boy jumps from a height of 4m onto a plate-form mounted on springs. As the

denpristay [2]

Answer:

c. 1600J

Explanation:

The loss in potential energy of the boy is given by:

$U=mg \Delta h$

where

m = 40 kg is the mass of the boy

g = 9.8 m/s^2 is the acceleration of gravity

$\Delta h = 4 m + 0.02 m = 4.02 m$ is the total change in the height of the boy (4 metres + 2 cm due to the compression of the spring)

Substituting, we find

$\Delta U = (40 kg)(9.8 m/s^2)(4.02 m) = 1577 J \sim 1600 J$

4 0

2 years ago