Ask question

Ask question

All categories

3 years ago

11

You walk 45 m to the north, then turn 90° to your right and walk another 45 m How far are you from where you originally started?

a.45 m b.41 m c.85 m d.64 m

1 answer:

docker41 [41]3 years ago

3 0

C^2=a^2+b^2
c^2=45^2+45^2
c^2=4050
c=63.64
c=64
The answer is D.

You might be interested in

If the sprinter accelerates at that rate for a distance of 15 m, and then maintains the velocity he has at that point for the re

ahrayia [7]

Answer:

The time for the entire race is 11.39 sec.

Explanation:

Given that,

Distance = 15 m

Remainder distance = 100 m

Suppose A sprinter begins a race with an acceleration of 3.4 m/s².

We need to calculate the time

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value in the equation

$15=0+\dfrac{1}{2}\times3.4\times t^2$

$t^2=\dfrac{30}{3.4}$

$t=2.97\ sec$

We need to calculate the final velocity of sprinter

Using equation of motion again

$v=u+at$

Put the value into the formula

$v=0+3.4\times2.97$

$v=10.09\ m/s$

We need to calculate the distance covers by sprinter

$d=100-15=85\ m$

The sprinter need to covers only 85 m.

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t'=\dfrac{85}{10.09}$

$t'=8.42\ sec$

We need to calculate the time for the entire race

$t''=t+t'$

Put the value into the formula

$t''=2.97+8.42$

$t''=11.39\ sec$

Hence, The time for the entire race is 11.39 sec.

4 0

2 years ago

Use the drop-down menus to complete each sentence.

qaws [65]

Answer:b

Explanation:

6 0

2 years ago

Question 6 Unsaved How many neutrons does element X have if its atomic number is 29 and its mass number is 84? Your Answer:

belka [17]

84-29=55 neutrons in nucleus

3 0

2 years ago

1. An object is moving at 4.66 m/s when it accelerates at 5.66 m/s2 for a period of 2.35 s. The distance it moves in this time i

lesya [120]

Answer:

1. Distance, S = 26.58 meters

2. Acceleration of gravity, g = 58.60 m/s²

Explanation:

1. Given the following data;

Initial velocity = 4.66 m/s

Acceleration = 5.66 m/s²

Time = 2.35 seconds

To find the distance travelled by the object, we would use the second equation of motion;

S = ut + ½at²

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

S = 4.66*2.35 + ½*5.66*2.35²

S = 10.951 + (2.83 * 5.5225)

S = 10.951 + 15.629

S = 26.58 meters

2. Given the following data;

Displacement (height) = 1.55 m

Time, t = 0.23 seconds

To find the acceleration of gravity, we would use the following formula;

Height = ½gt²

Substituting into the formula, we have;

1.55 = ½ * g * 0.23²

1.55 = ½ * g * 0.0529

1.55 = 0.02645g

Acceleration of gravity, g = 1.55/0.02645

Acceleration of gravity, g = 58.60 m/s²

5 0

2 years ago

A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are verti

NeX [460]

Answer:

a) T_A = $\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )$ , b) T_B = g [m₂ ( $\frac{x}{d} -1$ ) + m₁ ( $\frac{L}{ 2d} -1$ ) ]

c) x = $d - \frac{m_1}{m_2} \ \frac{L}{2d}$ , d) m₂ = m₁ ( $\frac{ L}{2d} -1$ )

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

∑ τ = 0

T_A d - W₂ x -W₁ L/2 = 0

T_A = $\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )$

b) the tension in string B

we write the expression of the translational equilibrium

∑ F = 0

T_A - W₂ - W₁ - T_B = 0

T_B = T_A -W₂ - W₁

T_ B = $\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )$ - g m₂ - g m₁

T_B = g [m₂ ( $\frac{x}{d} -1$ ) + m₁ ( $\frac{L}{ 2d} -1$ ) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

g m₂ (d -x) - g m₁ (0.5 L -d) + 0 = 0

m₂ (d-x) = m₁ (0.5 L- d)

m₂ x = m₂ d - m₁ (0.5 L- d)

x = $d - \frac{m_1}{m_2} \ \frac{L}{2d}$

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

0 = d - \frac{m_1}{m_2} \ \frac{L}{2d}

$\frac{m_1}{m_2} \ (0.5L -d) = d$

$\frac{m_1}{m_2} = \frac{ d}{0.5L-d}$

m₂ = m₁ $\frac{0.5 L -d}{d}$

m₂ = m₁ ( $\frac{ L}{2d} -1$ )

5 0

2 years ago