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3 years ago

5

Nathan is walking to the store and sees a snake slithering across the sidewalk. He jumps over it with an initial vertical veloci

ty of 10 ft/s. How long does it take for Nathan to land back on the ground

1 answer:

Travka [436]3 years ago

8 0

Answer:

$0.62\:\mathrm{s}$

Explanation:

Since the universal SI unit for velocity is meters/second, let's convert ft/s to m/s:

$10\:\mathrm{ft/s}=3.048\:\mathrm{m/s}$

We can use the following kinematics equation to solve this question:

$v_f=v_i+at$

What we know:

The initial velocity, $v_i$ , is $3.048\:\mathrm{m/s}$

(physics concept) The final velocity must be equal in magnitude but opposite in direction to the initial velocity ( $v_f=-3.048\:\mathrm{m/s}$ )
Acceleration, $a$ , is acceleration due to gravity at about $9.8\:\mathrm{m/s}$

Solving for $t$ :

$-3.048=3.048+(-9.8t),\\-6.096=-9.8t,\\t=\frac{-6.096}{-9.8}\approx \boxed{0.62\:\mathrm{s}}$

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Lana wanted to do an investigation to see what kind of birdseed the birds in her neighborhood like best. She purchased three of

Misha Larkins [42]

Answer:

The level of seed in each feeder

Explanation:

The independent variable is the variable which is the suspected cause of an observation, it is the variable that produce the effect observed in the dependent variable. The dependent variable is the variable that is measured

The independent variable is normally the x-value while the dependent variable is the y-value

In the question, Lana wants to find out the kind of birdseed that the neighborhood birds like (to eat) the most by feeding them different types of birdseed and measuring the level of the birdseed in the feeders at regular intervals

the independent variable is the bird seed types which Lana gives to the birds, to determine the type of bird seeds the birds like

The dependent variable is the level of the seed in the feeder which depends on the type of seeds the birds like.

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3 years ago

Describe the direction in which each object will accelerate. If the object won't accelerate, write "no acceleration." Explain wh

fgiga [73]

Answer:

No acceleration. All of the forces are acting equally on the paper air plane therfore It isn't going to move any where.

5 0

3 years ago

What eccentricity value results in a circular orbit?

Oxana [17]

Answer:

Zero

Explanation:

Given the equation of an ellipse:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

The eccentrity of an ellipse is given by:

$e=\sqrt{1-\frac{b^2}{a^2}}$

For a circle, we have

$a=b$

Therefore the eccentricity of a circle is

$e=\sqrt{1-\frac{1^2}{1^2}}=0$

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3 years ago

The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass

devlian [24]

Answer:

The velocity of the recoil is $v=1.001 \frac{m}{s}$

Explanation:

Kinetic Energy

$m_{bullet}*v_{bullet}=m_{gun}*v_{recoil}\\m_{gun}= 15.0kg+3.6kg$

The mass of th gun is the both mass the shotgun and the arm shoulder combination

$m_{bullet}=0.049kg\\v_{bullet}=380\frac{m}{s} \\m_{bullet}*v_{bullet}=m_{gun}*v_{recoil}\\0.049kg*380\frac{m}{s}=(15.kg+3.6kg)* v_{recoil}\\v_{recoil}=-\frac{18.62 kg \frac{m}{s} }{18.6 kg}\\ v_{recoil}=-1.0010 \frac{m}{s}$

The velocity is negative because is in opposite direction of the bullet

6 0

3 years ago

Read 2 more answers

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

3 years ago