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3 years ago

Assignments

Physics

1 answer:

sladkih [1.3K]3 years ago

4 0

Hello there!

I hope you and your family are staying safe and healthy during this unprecendented time.

A) What is the work done?

Answer: We need to use the formula

$w=-F_f(d)$

$w=-(35)(20)$

$w=-700J$

B) What is the work done on the cart by the gravitational force?

Alright, we know that the gravitional force is perpendicular to the diplacement. Therefore, we gonna use the following formula:

$w=Fdcos90$

$w=0$

C) What is the work done on the cart by the shopper?

This is the easier part, since we already know that the work done by the shopper is the same as the work done by the friction force

$W shopper + W friction = 0\\W shopper = W-friction \\W shopper = 700J$

D) Find the force the shopper exerts, using energy considerations.

$F_f+Fcos25=0\\-35+Fcos25=0\\F=38.6N$

E) What is the total work done?

You just need to add them:

$w=wshopper+wfriction\\w=0$

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If vector A =i+2j-k and vec A cross vec B =3i-j+5k. find vec B

postnew [5]

Let B = a i + b j + c k. Then the cross product of A = i + 2j - k with B is

A × B = ( i + 2j - k ) × ( a i + b j + c k )

A × B = a ( i × i ) + 2a ( j × i ) - a ( k × i )

… … … + b ( i × j ) + 2b ( j × j ) - b ( k × j )

… … … + c ( i × k ) + 2c ( j × k ) - c ( k × k )

A × B = 0 - 2a k - a j

… … … + b k + 0 + b i

… … … - c j + 2c i - 0

A × B = (b + 2c) i + (-a - c) j + (b - 2a) k

So we have

3 i - j + 5 k = (b + 2c) i + (c - a) j + (b - 2a) k

which gives us the system of equations,

{ b + 2c = 3

{ -a - c = -1

{ -2a + b = 5

Solve for a, b, and c.

• Eliminate c from the first two equations:

(b + 2c) + 2 (-a - c) = 3 + 2 (-1)

-2a + b = 1

But -2a + b = 5, and 5 ≠ 1, so there is no such vector B that satisfies the cross product!

8 0

2 years ago

A professional golfer hits a golf ball of mass 46 g with her 5-iron, and the ball first strikes the ground 155 m away. The ball

RideAnS [48]

Answer:

$C=2.32\times 10^{-4}\ Ns^2/m^2$

Explanation:

It is given that,

Mass of the golf ball, m = 46 g = 0.046 kg

Terminal speed of the ball, v = 44 m/s

The drag force, $F_r=Cv^2$

Where, C is the drag coefficient. At terminal speed, the weight of the ball is balanced by the drag force.

$Cv^2=mg$

$C=\dfrac{mg}{v^2}$

$C=\dfrac{0.046\times 9.8}{(44)^2}$

$C=2.32\times 10^{-4}\ Ns^2/m^2$

Hence, this is the required solution.

4 0

3 years ago

HELP MY GRADES-

belka [17]

If you count the number of seconds between the flash of lightning and the sound of thunder, and then divide by 5, you'll get the distance in miles to the lightning: 5 seconds = 1 mile, 15 seconds = 3 miles, 0 seconds = very close. Keep in mind that you should be in a safe place while counting.

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2 years ago

A 878-kg (1940 lb) dragster, starting from rest, attains a speed of 25.9 m/s (57.9 mph) in 0.62 s. (a) Find the average accelera

salantis [7]

Answer:

41.8m/s^2

Explanation:

Since the dragster starts from rest, initial velocity (u) = 0m/s, final velocity (v) = 25.9m/s, time (t) = 0.62s

From the equations of motion, v = u + at

a = (v - u)/t = (25.9 - 0)/0.62 = 25.9/0.62 = 41.8m/s^2

7 0

2 years ago

What is the safety rules?

Black_prince [1.1K]

Answer:

Quick question do you mean what are some safety rules

Explanation:

Crosswalk, Stop sign,

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3 years ago

Read 2 more answers