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3 years ago

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While Bob is demonstrating the gravitational force on falling objects to his class, he drops an 1.0 lb bag of feathers from the

top of the science building. Determine the distance the bag has traveled after falling for 1.5 seconds assuming it has reach free fall and given the gravitational acceleration of 9.8 m/sec2.
Distance = velocity x time

A)
7.4 m

B)
11 m

C)
15 m

D)
22 m

1 answer:

Sunny_sXe [5.5K]3 years ago

6 0

Answer:

Option B, 11 m is the correct answer.

Explanation:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case u = 0 m/s , t = 1.5 seconds , a = acceleration due to gravity = $9.8m/s^2$ , we need to find displacement

Substituting

$s= 0*1.5+\frac{1}{2}*9.8*1.5^2\\ \\ s=11.025m$

So option B, 11 m is the correct answer.

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Light passes from a material A, which has an index of refraction of 4/3, into material B, which has an index of refraction of 5/

katrin2010 [14]

Answer:

(b) 32.2°

Explanation:

Using Snell's law as:

$n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence ( 30.0° )

${\theta_r}$ is the angle of refraction ( ? )

${n_r}$ is the refractive index of the refraction medium (Material B, n=5 / 4)

${n_i}$ is the refractive index of the incidence medium (Material A, n=4 / 3)

Hence,

${\frac {4}{3}}\times {sin30.0^0}={\frac {5}{4}}\times{sin\theta_r}$

Angle of refraction = $sin^{-1}0.5333$ = 32.2°

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4 years ago

. Maria walked 1.5 miles south to her house in 0.5 hours. What is her speed in miles per hour?

Simora [160]

1) 3 miles/Hour

The speed is defined as the distance covered divided by the time taken:

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

where

d = 1.5 mi is the distance

t = 0.5 h is the time taken

Substituting,

$v=\frac{1.5}{0.5}=3 mi/h$

2) 1.34 m/s south

Velocity, instead, is a vector, so it has both a magnitude and a direction. We have:

$d=1.5 mi \cdot 1609 m/mi = 2414 m$ is the displacement in meters

$t=0.5 h \cdot 3600 s/h =1800 s$ is the time taken in seconds

Substituting,

$v=\frac{2414 m}{1800 s}=1.34 m/s$

And the direction of the velocity is the same as the displacement, so it is south.

6 0

4 years ago

When an ice cube is placed in a glass of warm water, how are the signs and values q for the ice and the warm water related, assu

Valentin [98]

Answer:

q(ice) = -q(warm water)

Explanation:

Ice removes energy if ice is put in warm water and ice retains the very same amount of energy. Thus the water temperature heat sign is negative, and the ice heat sign is positive.

3 0

3 years ago

A 80 kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 25 degrees hill.

zloy xaker [14]

Answer:

$P=28.085\,hp$

Explanation:

Given that:

mass of 1 skier, $m=80kg$

inclination of hill, $\theta=25^{\circ}$

length of inclined slope, $l=220m$

time taken to reach the top of hill, $t=2.3 min= 138 s$

coefficient of friction, $\mu=0.15$

<em>Now, force normal to the inclined plane:</em>

$F_N=m.g.cos\theta$

$F_N=80\times 9.8\times cos25^{\circ}$

$F_N=710.54\,N$

<em>Frictional force:</em>

$f=\mu.F_N$

$f=0.15\times 710.54$

$f=106.58\,N$

<em>The component of weight along the inclined plane:</em>

$W_l=m.g.sin\theta$

$W_l=80\times 9.8\times sin25^{\circ}$

$W_l=331.33\,N$

<em>Now the total force required along the inclination to move at the top of hill:</em>

$F=f+W_l$

$F=106.58+331.33$

$F=437.91\,N$

<em>Hence the work done:</em>

$W=F.l$

$W=437.91\times 220$

$W=96340.80\,J$

<em>Now power:</em>

$P=\frac{W}{t}$

$P=\frac{96340.80}{138}$

$P=698.12\,W$

<u>So, power required for 30 such bodies:</u>

$P=30\times 698.12$

$P=20943.65\,W$

$P=\frac{20943.65}{745.7}$

$P=28.085\,hp$

8 0

3 years ago

Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d

Step2247 [10]

Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that $F = k \frac{Q_1Q_2}{d^2}$ so if we double the charge on $Q_1$ and double the distance to $(2d)$ we plug these into the equation to find

<span> $F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}$ </span>

So we see the new force is exactly 1/2 of the old force so your answer should be $\frac{1}{2}F$ if I can remember my physics correctly.

9 0

4 years ago

Read 2 more answers