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4 years ago

5

Lincoln weighs 400 newtons. What’s his mass rounded to the nearest kilogram? Assume that acceleration due to gravity is 9.8 N/kg

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2 answers:

Yanka [14]4 years ago

6 0

Weight equals mass times gravitational acceleration=400N, so mass=400/9.8=41kg approx.

Serggg [28]4 years ago

6 0

Answer:

41 k

Explanation:

Use the formula for weight: W = m × g. Rearrange the formula to solve for m:

m = W ÷ g.

Substitute W = 400 newtons and g = 9.8 N/kg into the formula. Then divide:

m = 400 N ÷ 9.8 N/kg

40.8 kg.

Rounded to the nearest kilogram, Lincoln has a mass of 41 kilograms.

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A ship is towed through a narrow channel by applying forces to three ropes attached to its bow. Determine the magnitude and orie

Y_Kistochka [10]

This question is solved using an available similar problem as data provided for the forces was not given.

Repeat the same steps outlined for your problem.

Regards.

Answer:

F = 1.598 KN , Q = 90 degree (+ y-axis)

Explanation:

Sum of Forces in x-direction to the left (+)

2 cos (30) + 3cos (60) + F*cos (Q) = F_a ..... 1

Sum of Forces in y-direction to the up (+)

2 sin (30) + F*sin (Q) - 3 sin (60) ...... 2

Using Eq 2 and solve:

F*sin (Q) = 1.598 KN

F_min when sin (Q) is max, max possible value of sin(Q) = 1 @ Q = 90 degrees.

Hence,

F_min = 1.598 KN

Using Eq 1 @ Q = 90 degrees and F = 1.598 KN:

F_a = 2 cos (30) + 3cos (60) = 3.2 KN

6 0

3 years ago

A 38.5kg man is in an elevator accelerating downward. A normal force of 343n pushes up on him. what is his acceleration?

alexira [117]

Answer:

<h3>The answer is 8.91 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{343}{38.5} = \frac{98}{11} \\ = 8.909090...$

We have the final answer as

<h3>8.91 m/s²</h3>

Hope this helps you

4 0

3 years ago

If the direction of the position is north and the direction of the velocity is up, then what is the direction of the angular mom

Kay [80]

Answer:

the direction of angular momentum = EAST

Explanation:

given

Direction of position = r = north

Direction of velocity = v = up

angular momentum = L = m(r x v)

where m is the mass, r is the radius, v is the velocity

utilizing the right hand rule, the right finger heading towards the course of position vector and curl them toward direction of velocity, at that point stretch thumb will show the bearing of the angular momentum.

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6 0

3 years ago

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to

White raven [17]

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^{-1}$

speed of the swimmer with respect to water, $v=4\ ft.s^{-1}$

<u>Now the resultant of the two velocities perpendicular to each other:</u>

$v_r=\sqrt{v^2+v_s^2}$

$v_r=\sqrt{4^2+8^2}$

$v_r=8.9442\ ft.s^{-1}$

<u>Now the angle of the resultant velocity form the vertical:</u>

$\tan\beta=\frac{v_s}{v}$

$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d\times \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=\sqrt{d^2-w^2}$

$\Delta s=\sqrt{1118.034^2-500^2}$

$\Delta s=1000\ ft$

2)

On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

$v'=v.\cos37$

$v'=4\times \cos37$

$v'=3.1945\ ft.s^{-1}$

<u>Now the net effective speed of stream sweeping the swimmer:</u>

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

<u>The component of swimmer's velocity heading directly towards the opposite bank:</u>

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^{-1}$

<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

$\phi=63.39^{\circ}$

Now let the distance swam in this direction be d'.

$d'\times \cos\phi=w$

$d'=\frac{500}{\cos63.39}$

$d'=1116.344\ ft$

<u>Now the distance swept downstream:</u>

$\Delta s'=\sqrt{d'^2-w^2}$

$\Delta s'=\sqrt{1116.344^2-500^2}$

$\Delta s'=998.11\ ft.s^{-1}$

3)

Time taken in crossing the rive in case 1:

$t=\frac{d}{v_r}$

$t=\frac{1118.034}{8.9442}$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

7 0

4 years ago

A rookie quarterback throws a football with an initial upward velocity component of 17.0 m/s and a horizontal velocity component

vitfil [10]

Answer:

(a) 1.73 s

(b) 14.75 m

(c) 3.36 s

(d) double

(e) 63.32 m

Explanation:

Vertical component of initial velocity, uy = 17 m/s

Horizontal component of initial velocity, ux = 18.3 m/s

(A) At highest point of trajectory, the vertical component of velocity is zero. Let the time taken is t.

Use first equation of motion in vertical direction

vy = uy - gt

0 = 17 - 9.8 t

t = 1.73 seconds

(B) Let the highest point is at height h.

Use III equation of motion in vertical direction

$v^{2}=u^{2}-2gh$

0 = 17 x 17 - 2 x 9.8 x h

h = 14.75 m

(C) The time taken by the ball to return to original level is T.

Use second equation of motion i vertical direction.

$h = ut + 0.5at^2$

h = 0 , u = 17 m/s

0 = 17 t - 0.5 x 9.8 t^2

t = 3.46 second

(D) It is the double of time calculated in part A

(E) Horizontal distance = horizontal velocity x total time

d = 18.3 x 3.46 = 63.32 m

8 0

3 years ago