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2 years ago

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This athlete loads a sled to a combined mass of 90.0 kg, and drags the sled with a force of 258 N at an angle of 15.0° across a

surface with a coefficient of kinetic friction of 0.250. What is the approximate acceleration of the sled?
Group of answer choices

1 answer:

Semenov [28]2 years ago

4 0

Answer:

The approximate acceleration of the sled is 0.319 m/s²

Explanation:

Given;

mass of the sled, m = 90 kg

horizontal applied force, f = 258 N

angle of inclination of the force, θ = 15°

coefficient of kinetic friction of 0.250, μ = 0.25

The net vertical force on the sled is given by;

Fcosθ - μmg = ma

258Cos15 - (0.25 x 90 x 9.8) = ma

249.2 - 220.5 = ma

28.7 = ma

a = 28.7 /m

a = 28.7 / 90

a = 0.319 m/s²

Therefore, the approximate acceleration of the sled is 0.319 m/s²

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A drag racer accelerates from rest at an average rate of +13.2 mls for a distance of 100. m. The driver coasts for 0.5 then uses

gtnhenbr [62]

Complete Question:

A drag racer accelerates from rest at an average rate of +13.2 m/s² for a distance of 100. m. The driver coasts for 0.5 s then uses the brakes and parachute to decelerate until the end of the track. If the total length of the track is 180 m, what minimum deceleration rate must the racer have in order to stop prior to the the end of the track?

Answer:

-31.92 m/s²

Explanation:

The drag races do a retiling uniform variated movement. There are 3 steps in the movement, first, it accelerates from rest until 100 m, second it coasts to 0.5 s, and then it decelerates. So, let's analyze each one of the steps:

Step 1

The initial velocity is v0 = 0 (because it was at rest), the acceleration is +13.2 m/s², and the distance ΔS = 100.0 m, so the final velocity, v, is:

v² = v0² + 2aΔS

v² = 2*13.2*100

v² = 2640

v = √2640

v = 51.38 m/s

Step 2

Know it's initial velocity is 51.38 m/s, it take 0.5s, and has the same acceleration, so, after 0.5 s, the velocity will be:

v = v0 + at

v = 51.38 + 13.2*0.5

v = 57.98 m/s

Thus, the distance it travels is:

v² = v0² + 2aΔS

57.98² = 51.38² + 2*13.2*ΔS

3361.6804 = 2639.9044 + 26.4ΔS

26.4ΔS = 721.776

ΔS = 27.34 m

Step 3

The initial velocity of the drag racer is 57.98 m/s, and it travels the final distance of the track: 180 - 100 - 27.34 = 52.66 m. So, when it stops, its final velocity will be 0. The minimum deceleration must be the one that it would stop at the end of the track (less than that it would cross the final track):

v² = v0² + 2aΔS

0 = 57.98² + 2a*52.66

-105.32a = 3361.6804

a = - 31.92 m/s²

4 0

2 years ago

When jumping, a flea accelerates at an astounding 1000 m/s2 but over the very short distance of 0.50 mm. If a flea jumps straigh

Nadusha1986 [10]

Answer:

The flea reaches a height of 51 mm.

Explanation:

Hi there!

The equations of height and velocity of the flea are the following:

During the jump:

h = h0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

While in free fall:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the flea at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration of the flea due to the jump.

v = velocity of the flea at time t.

g = acceleration due to gravity.

First, let's calculate how much time it takes the flea to reach a height of 0.0005 m. With that time, we can calculate the speed reached by the flea during the jump:

h = h0 + v0 · t + 1/2 · a · t²

If we place the origin of the frame of reference on the ground, then, h0 = 0. Since the flea is initially at rest, v0 = 0. Then:

h = 1/2 · a · t²

We have to find the value of t for which h = 0.0005 m:

0.0005 m = 1/2 · 1000 m/s² · t²

0.0005 m / 500 m/s² = t²

t = 0.001 s

Now, let's find the velocity reached in that time:

v = v0 + a · t (v0 = 0)

v = a · t

v = 1000 m/s² · 0.001 s

v = 1.00 m/s

When the flea is at a height of 0.50 mm, its velocity is 1.00 m/s. This initial velocity will start to decrease due to the downward acceleration of gravity. When the velocity is zero, the flea will be at the maximum height. Using the equation of velocity, let's find the time at which the flea is at the maximum height (v = 0):

v = v0 + g · t

At the maximum height, v = 0:

0 m/s = 1.00 m/s - 9.81 m/s² · t

-1.00 m/s / -9.81 m/s² = t

t = 0.102 s

Now, let's find the height reached by the flea in that time:

h = h0 + v0 · t + 1/2 · g · t²

h = 0.0005 m + 1.00 m/s · 0.102 s - 1/2 · 9.81 m/s² · (0.102 s)²

h = 0.051 m

The flea reaches a height of 51 mm.

5 0

2 years ago

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$

Substitute Eq 2 into Eq 1

$0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}$

Using Eq 1

$0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m$

7 0

2 years ago

A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur

Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0

2 years ago

A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold

LUCKY_DIMON [66]

Answer:

(a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

$Q=\dfrac{V}{t}$

Put the value into the formula

$Q=\dfrac{405}{90}$

$Q=4.5\ cm^3/s$

(a). We need to calculate the coefficient of permeability

Using formula of coefficient of permeability

$Q=kiA$

$k=\dfrac{Q}{iA}$

$k=\dfrac{Ql}{Ah}$

Where, Q=discharge

l = length

A = cross section area

h=constant head causing flow

Put the value into the formula

$k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}$

$k=8.6\times10^{-3}\ cm/s$

The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(c). We need to calculate the discharge velocity during the test

Using formula of discharge velocity

$v=ki$

$v=\dfrac{kh}{l}$

Put the value into the formula

$v=\dfrac{8.6\times10^{-3}\times38}{17.5}$

$v=0.0187\ cm/s$

The discharge velocity during the test is 0.0187 cm/s.

(b). We need to calculate the volume of solid in the ample

Using formula of volume

$V_{s}=\dfrac{M_{s}}{V_{s}}$

Put the value into the formula

$V_{s}=\dfrac{4950\times10^{-3}}{2710}$

$V_{s}=1826.56\ cm^3$

We need to calculate the volume of the soil specimen

Using formula of volume

$V=A\times L$

Put the value into the formula

$V=\dfrac{\pi(17.5)^2}{4}\times17.5$

$V=4209.24\ cm^3$

We need to calculate the volume of the voids

$V_{v}=V-V_{s}$

Put the value into the formula

$V_{v}=4209.24-1826.56$

$V_{v}=2382.68\ cm^3$

We need to calculate the seepage velocity

Using formula of velocity

$Av=A_{v}v_{s}$

$v_{s}=\dfrac{Av}{A_{v}}$

$v_{s}=\dfrac{V}{V_{v}}\times v$

Put the value into the formula

$v_{s}=\dfrac{4209.24}{2382.68}\times0.0187$

$v_{s}=0.0330\ cm/s$

The seepage velocity is 0.0330 cm/s.

Hence, (a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

8 0

2 years ago