\Which are consumers within a spruce-fir forest ecosystem?

A box falls off of a tailgate and slides along the street for a distance of 62.5 m. Friction slows the box at –5.0 m/s2. At what

Mila [183]

Answer:

25 m/s

Explanation:

This question can be solved using equation of motion

$v^2 = u^2 + 2as$

where

v is the final velocity

u is the initial velocity

s is the distance covered while moving from initial to final velocity

a is the acceleration

_____________________________________________

Given

box moved for distance of 62.5 m

Friction slows the box at –5.0 m/s2----> this statement means that there is deceleration , speed of truck decreases by 5 m/s in every second until the box comes to rest. Friction causes this deceleration.

thus in this problem

a = -5.0 m/s2

V = 0 as body came to rest due to friction deceleration

u the initial velocity we have to find

the initial velocity of box will be the same as speed of truck, as the box was in the truck and hence box will pick the speed of truck.

so if we find speed of box, we will be able get sped of truck as well.

using equation of motion

$v^2 = u^2 + 2as\\0^2 = u^2 + 2*-5* 62.5\\0 = u^2 - 625\\u^2 = 625\\\sqrt{u^2} = \sqrt{625} \\u = 25$

Thus, initial speed with the truck was travelling was 25 m/s.

3 0

2 years ago

Two 1.9 kg masses are 1.1 m apart (center to center) on a frictionless table. Each has + 9.6 μC of charge.

Anni [7]

Answer:

F= 0.6 N

Explanation:

Fe(electrical force)= $k q1q2/r^2$

$k=9*10^9\\q1=-9.6 *10^-6 C\\q2= -9.6*10^-6 C\\r= 1.1m$

So,

$F=\frac{9*10^9*-9.6 *10^-6 C* -9.6*10^-6 C}{1.1^2}$

$F= 0.6 N$

5 0

2 years ago

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0

2 years ago

FM radio ________________. a. had a somewhat shorter range than AM radio, but better sound quality. b. was widely adopted in the

svetlana [45]

Answer:

(A) FM Radio had a somewhat shorter ranger than AM radio, but better sound quality.

Explanation:

FM Radio was invented in 1933 by Edwin Armstrong who was an American engineer. FM stands for frequency modulation and AM stands for Amplitude Modulation.

FM is used for most broadcasts of music and FM radio stations use a very high-frequency range of radio frequencies.

In FM Radio, the sound is transmitted through changes in frequency. Both FM and AM radio signals experience frequent change in amplitude, they are far less noticeable on FM.

When switching between stations, FM antenna is alternating between different frequencies, and not amplitudes and this produces a much clearer sound and allows for smoother transitions with little to no audible static.

FM signals can be interfered by barriers and this could affect the signal strength. FM Radio signals are more clearer in a mountainous area that has no barrier.

AM radio was able to carry signals farther than AM radio.

6 0

3 years ago

A bolt is dropped from a bridge under construction, falling 96 m to the valley below the bridge. (a) How much time does it take

irakobra [83]

Answer:

a)It takes the bolt 0.25 s to pass the last 11% of the fall.

b)When the bolt begins to fall the last 11% of the fall its velocity is -41.2 m/s.

c)The velocity of the bolt just before it reaches the ground is -43.6 m/s

Explanation:

Hi there!

a) Let´s calculate how much distance it is the last 11% of the fall:

96 m · 0.11 = 10.56 m

So, we have to find how much time it takes the bolt to pass from a height of 10.56 m to the ground.

First, let´s calculate how much time it takes the bolt to reach a height of 10.56 m. For that we can use this equation:

h = h0 + v0 · t + 1/2 · g · t²

Where:

h = height of the bolt at a time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity.

If we consider the ground as the origin of the frame of reference, then h0 = 96 m. Since the bolt is dropped, the initial velocity is zero (v0 = 0). Then, the equation gets reduce to this:

h = h0 + 1/2 · g · t²

We have to find at which time h = 10.56 m.

10.56 m = 96 m - 1/2 · 9.8 m/s² · t²

Solving for t:

√(-2 · (10.56 m - 96 m) / 9.8 m/s²) = t

t = 4.2 s

Now that we have the time at which the bolt is located at 10.56 m above the ground, we can calculate the velocity of the bolt at that time.

The equation of velocity (v) of the bolt is the following:

v = v0 + g · t

at t = 4.2 s.

v = 0 - 9.8 m/s² · 4.2 s

v = -41.2 m/s

When the bolt begins to fall the last 11% of the fall its velocity is -41.2 m/s.

Now, we can calculate how much time it takes to fall the last 10.56 m.

The initial velocity of the bolt will be the velocity at h = 10.56 m. The initial height will be 10.56 m.

h = h0 + v0 · t + 1/2 · g · t²

We have to find the time at which h = 0 (the bolt hits the ground)

0 = 10.56 m - 41.2 m/s · t - 1/2 · 9.8 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 0.25 s (the other solution of the quadratic equation is negative and thus discarded).

It takes the bolt 0.25 s to pass the last 11% of the fall.

Now, let´s calculate the velocity of the bolt when it reaches the ground:

v = v0 + g · t

v = -41.2 m/s - 9.8 m/s² · 0.25 s

v = -43.6 m/s

The velocity of the bolt just before it reaches the bolt is -43.6 m/s

6 0

2 years ago