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3 years ago

15

If a hockey puck is made to slide along a straight line on a completely frictionless surface, which statement about the motion o

f the hockey puck is true?
The hockey puck will come to a halt immediately.

The hockey puck will continue moving with constant velocity along a straight line.

The hockey puck will accelerate and continue moving in straight line.

The hockey puck will continue moving in random direction.

2 answers:

77julia77 [94]3 years ago

8 0

C. the hockey puck will accelerate and continue moving in a straight line

Llana [10]3 years ago

7 0

Answer:

The hockey puck will continue moving with constant velocity along a straight line.

Explanation:

As per Newton's first law we know that when there is no net force on an object then the velocity of object will always remains constant

Or we can say that the acceleration of an object will be zero when the force applied on the object is zero.

As per Newton's law we know

$F = ma$

so we will have

a = 0

so object will continue in straight line with same uniform speed without any change in magnitude as well as any change in direction

so correct answer will be

The hockey puck will continue moving with constant velocity along a straight line.

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A diverging lens has a focal length of 18.0 cm. An insect is placed 7.00 cm in front of the lens. What is the magnification

musickatia [10]

Diverging lens=concave lens

The image would be after the first focal point but before the lens and it will be smaller than the object. Look at my picture for reference. Please let me know if this helped you!

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2 years ago

Find a unit vector normal to the plane containing Bold u equals 2 Bold i minus Bold j minus 3 Bold k and Bold v equals negative

Nastasia [14]

Answer:

$\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}$

Explanation:

It is given that,

$\vec{u}=2i-j-3k$

$\vec{v}=-3i+j-2k$

Taking the cross product of v and v such that,

$\vec{w}=u\times v$

$\vec{w}=(2i-j-3k)\times (-3i+j-2k)$

$\vec{w}=5i+13j-k$

$|w|=\sqrt{5^2+13^2(-1)^2}$

|w| = 13.92

Let $\hat{w}$ is the unit vector normal to the plane containing u and v. So,

$\hat{w}=\dfrac{\vec{w}}{|w|}$

$\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}$

Hence, this is the required solution.

7 0

3 years ago

An object has a mass of 7g and a volume of 14cm. What is the density

Andreyy89

Density = mass / volume = 7/14 = 0.5!g/cm^3

8 0

3 years ago

A steel aircraft carrier is 370 m long when moving through the icy North Atlantic at a temperature of 2.0 °C. By how much does t

34kurt

Answer:

The carrier lengthen is 0.08436 m.

Explanation:

Given that,

Length = 370 m

Initial temperature = 2.0°C

Final temperature = 21°C

We need to calculate the change temperature

Using formula of change of temperature

$\Delta T=T_{f}-T_{i}$

$\Delta T=21-2.0$

$\Delta T=19^{\circ}C$

We need to calculate the carrier lengthen

Using formula of length

$\Delta L=\alpha_{steel}\times L_{0}\times\Delta T$

Put the value into the formula

$\Delta L=1.2\times10^{-5}\times370\times19$

$\Delta L=0.08436\ m$

Hence, The carrier lengthen is 0.08436 m.

8 0

3 years ago

Find the direction and magnitude of the net force exerted on the point charge q3 in the figure. Let q= +2.4 μC and d= 33cm.

kobusy [5.1K]

With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees

ELECTRIC FORCE (F)

F = $\frac{KQq}{d^{2} }$

Where K = 9 x $10^{9}$ N $m^{2}$ / $C^{2}$

The distance between $q_{1}$ and $q_{3}$ can be calculated by using Pythagoras theorem.

d = $\sqrt{33^{2} + 33^{2} }$

d = 46.7 cm = 0.467 m

For force $F_{1}$ , substitute all the parameters into the formula above

$F_{1}$ = (9 x $10^{9}$ x 3 x 1)/ $0.467^{2}$

$F_{1}$ = 2.7 x $10^{10}$ /0.218

$F_{1}$ = 1.24 x $10^{11}$ N

For force $F_{4}$ , substitute all the parameters into the formula above

$F_{4}$ = (9 x $10^{9}$ x 3 x 4)/ $0.33^{2}$

$F_{4}$ = 1.08 x $10^{11}$ /0.1089

$F_{4}$ = 9.92 x $10^{11}$ N

For force $F_{2}$ , substitute all the parameters into the formula above

$F_{2}$ = (9 x $10^{9}$ x 3 x 2)/ $0.33^{2}$

$F_{2}$ = 5.4 x $10^{10}$ /0.1089

$F_{2}$ = 4.96 x $10^{11}$ N

Summation of forces on Y component will be

$F_{y}$ = $F_{4}$ - $F_{1}$ Sin 45

$F_{y}$ = 9.92 x $10^{11}$ - 1.24 x $10^{11}$ Sin 45

$F_{y}$ = 9.04 x $10^{11}$ N

Summation of forces on X component will be

$F_{x}$ = $F_{2}$ - $F_{1}$ Cos 45

$F_{x}$ = 4.96 x $10^{11}$ - 1.24 x $10^{11}$ Sin 45

$F_{x}$ = 4.08 x $10^{11}$ N

Net Force = $\sqrt{F_{x} ^{2} + F_{y} ^{2} } }$

Net force = $\sqrt{(4.08*10^{11}) ^{2} + (9.04*10^{11}) ^{2} }$

Net force = 9.9 x $10^{11}$ N

The direction will be

Tan ∅ = $F_{y}$ / $F_{x}$

Tan ∅ = 9.04 x $10^{11}$ / 4.08 x $10^{11}$

Tan ∅ = 2.216

∅ = $Tan^{-1}$ (2.216)

∅ = 65.7 degrees

Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees approximately.

Learn more about electric Force here: brainly.com/question/4053816

8 0

2 years ago

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