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3 years ago

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Physics

1 answer:

slamgirl [31]3 years ago

8 0

Answer:

Speed at which it will reach the ground is given as

$v_f = 46.8 m/s$

Total time for which it will remain in air is given as

t = 6.3 s

Explanation:

As we know that the object is projected upwards with speed

$v_i = 15 m/s$

$g = - 9.81 m/s^2$

now when it will reach the ground then we have

$y = v_y t + \frac{1}{2} at^2$

so we have

$-100 = 15 t - \frac{1}{2}(-9.81) t^2$

$4.905 t^2 - 15 t - 100 = 0$

so we have

$t = 6.3 s$

Now speed of the object when it reaches the ground is given as

$v_f = v_i + at$

$v_f = -15 + (9.81)(6.3)$

$v_f = 46.8 m/s$

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The 64.5-kg climber in is supported in the “chimney” by the friction forces exerted on his shoes and back. The static coefficien

kobusy [5.1K]

Answer:

The minimum force the climber must exert is about 439N.

Explanation:

We use the relationship between friction and normal force to answer this question:

$F_{friction} = \mu_{static} \cdot F_{normal}\implies F_{normal}=\frac{F_{friction}}{\mu_{static}}$

We are given the static coefficients of friction but need to determine the friction force. To do that we consider the totality of forces acting on this hapless gentleman stuck in a chimney. There is the gravity acting downward (+), then there are two friction forces acting upward (-), namely through his shoes and his back. The horizontal force exerted by the climber on both walls of the chimney is the same and is met with equally opposing normal force. Since the climber is not falling the net force in the vertical direction is zero:

$F_{net} = 0 = F_g - F_{shoes}-F_{back}= mg - \mu_{shoes}F_{norm}-\mu_{back}F_{norm}\\F_{norm}=\frac{mg}{\mu_{shoes}+\mu_{back}}=\frac{64.5kg\cdot 9.8\frac{m}{s^2}}{0.8+0.64}\approx 438.96N\\$

The normal force in this equilibrium is about 439N and because we are told that the static friction forces are both at their maximum, this value is at the same time the <em>minimum</em> force needed for the climber to avoid starting slipping down the chimney.

7 0

3 years ago

What distance will a vehicle travel before coming to a complete stop from a speed of 70 mph, (a) When the vehicle is traveling o

OLga [1]

Answer:

(a), The SSD will be 723.9 ft.

(b-1), The SSD will be 620.2 ft.

(b-2), The SSD will be $723.91>SSD>620.2$

(c), The SSD will be 910.5 ft.

Explanation:

Given that,

Speed = 70 mph

Suppose, a perception reaction time of 2.5 sec and the coefficient of friction is 0.35

We need to calculate the stopping sight distance

Using formula of SSD

$SSD=1.47\times v\times t+\dfrac{v^2}{30\times(f\pm g)}$

Where, v = speed of vehicle

t = perception reaction time

f = coefficient of friction

g = gradient of road

(a). If the gradient of road is zero.

Then, the stopping sight distance will be

$SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35)}$

$SSD=723.9\ ft$

(b-1). If the gradient of road is 0.1

Then, the stopping sight distance will be

$SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35+0.1)}$

$SSD=620.2\ ft$

(b-2). If the grade continuously decrease then the SSD will be increase.

But if the grade is increase then the SSD will be decrease and for flat grade the SSD will be more.

So, The SSD will be $723.91>SSD>620.2$

(c). When the vehicle is traveling downhill on a roadway of constant grade then the vehicle take will be more SSD

So, The SSD will be

$SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35-0.1)}$

$SSD=910.5\ ft$

Hence, (a), The SSD will be 723.9 ft.

(b-1), The SSD will be 620.2 ft.

(b-2), The SSD will be $723.91>SSD>620.2$

(c), The SSD will be 910.5 ft.

7 0

3 years ago

PLS HELP!!

stealth61 [152]

Answer:

Explanation:

The iris controls the amount of light that enters the eye by opening and closing the pupil. The iris uses muscles to change the size of the pupil. These muscles can control the amount of light entering the eye by making the pupil larger (dilated) or smaller (constricted).

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2 years ago

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An object moving north with an initial velocity of 14 m/s accelerates 5 m/s2 for 20 seconds. What is the final velocity of the o

lisabon 2012 [21]

Answer:

option C

Explanation:

Final velocity of the object is 114 m / s. Hence, final velocity of the object is 114 m / s.

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3 years ago

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A toaster uses 6700 joules of energy in 45 seconds to toast to a piece of bread what is power of the oven

just olya [345]

P= w/t and W= Work
In this case, W= 6,700j, and T= 45 seconds
Power is the ratio of work per unit time. When you perform a work in a given span of time, the ratio of work performed with respect to time is Called Power.
si unit for Power is Watt (W)
so, P= 6,700/45
P= 148
Final answer is P=148

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3 years ago