Ask question

Ask question

All categories

tamaranim1 [39]

2 years ago

11

The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope. The sl

eds move on an icy surface. The dogs exert a 240-N force on the rope attached to the front sled. The front rope pulls horizontally.
Part A
Find the acceleration of the sleds.
Express your answer to two significant figures and include the appropriate units.
Part B
Find the force the rope between the sleds exerts on each sled.
Express your answer to two significant figures and include the appropriate units.

2 answers:

azamat2 years ago

7 0

Answer:

Part a)

$a = 2.4 m/s^2$

Part b)

$F = 120 N$

Explanation:

Two sleds are connected by a rope

mass of each sled is given as

$m = 100 kg$

now we know that dog exert pulling force on the rope connected to first sled

$F = 240 N$

Part a)

By newton's first law we know that

$F = (m + m)a$

$240 = (100 + 100)a$

$a = \frac{240}{100}$

$a = 2.4 m/s^2$

Part b)

As we know that force between two ropes will pull the sled behind

so we will have

$F = ma$

$F = 100(1.20)$

$F = 120 N$

loris [4]2 years ago

5 0

Answer:

Answer:

(a) acceleration a = 1.2m/s2

(b) force F = 100N on each sled

Explanation:

The explanation is in the attached file below

You might be interested in

A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t

Semenov [28]

Answer:

a. $I=2.77x10^{-8} kg*m^2$

b. $K=4.37 x10^{-6} N*m$

Explanation:

The inertia can be find using

a.

$I = m*r^2$

$m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg$

$r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m$

$I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2$

$I=2.77x10^{-8} kg*m^2$

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

$T=2\pi *\sqrt{\frac{I}{K}}$

Solve to K'

$K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}$

$K=4.37 x10^{-6} N*m$

3 0

2 years ago

The air surrounding an airplane in flight exerts a drag force that acts opposite to the airplane's motion. When an Airbus A380 i

Rainbow [258]

Answer:

$4.32\cdot 10^5 hp$ , $3.22\cdot 10^8 W$

Explanation:

The jet is flying at constant velocity: this means that its acceleration is zero, so the net force acting on the jet is also zero.

Therefore, we can write:

$4F_T - F_d = 0$

where

$F_T = 322,000 N$ is the thrust force generated by each engine of the jet

$F_d$ is the drag force

Solving for Fd,

$F_d = 4 F_T = 4(322,000)=1.288\cdot 10^6 N$

The velocity of the jet is

$v=250 m/s$

So, the rate at which the drag force does work (which is the power) is

$P=F_d v$

and substituting

$F_d = 1.288\cdot 10^6 N\\v = 250 m/s$

we find

$P=(1.288\cdot 10^6)(250)=3.22\cdot 10^8 W$

Converting into horsepower,

$P=\frac{3.22\cdot 10^8}{746}=4.32\cdot 10^5 hp$

4 0

2 years ago

A worker stands still on a roof sloped at an angle of 35° above the horizontal. He is prevented from slipping by static friction

aleksley [76]

Answer:

99.63 kg

Explanation:

From the force diagram

N = normal force on the worker from the surface of the roof

f = static frictional force = 560 N

θ = angle of the slope = 35

m = mass of the worker

W = weight of the worker = mg

W Cosθ = Component of the weight of worker perpendicular to the surface of roof

W Sinθ = Component of the weight of worker parallel to the surface of roof

From the force diagram, for the worker not to slip, force equation must be

W Sinθ = f

mg Sinθ = f

m (9.8) Sin35 = 560

m = 99.63 kg

5 0

2 years ago

A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mir

Semmy [17]

Answer:

The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

Explanation:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{0.24}=\dfrac{1}{0.25}+\dfrac{1}{u}$

$\dfrac{1}{u}=\dfrac{1}{0.24}-\dfrac{1}{0.25}$

$\dfrac{1}{u}=\dfrac{1}{6}$

$u=6\ m$

We need to calculate the magnification

Using formula of magnification

$m=-\dfrac{v}{u}$

Put the value into the formula

$m=-\dfrac{0.24}{-6}$

$m=0.04$

We need to calculate the height of the object

Using formula of magnification

$m=\dfrac{h'}{h}$

$h=\dfrac{0.080}{0.04}$

$h=2\ m$

A convex mirror produce a virtual and upright image behind the mirror.

Hence, The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

6 0

2 years ago

Read 2 more answers

You are on vacation in San Francisco and decide to take a cable car to see the city. A 5800-kgkg cable car goes 260 mm up a hill

Stella [2.4K]

Answer:

$4.325\times10^6J$

Explanation:

Mass of the cable car, m = 5800 kg

It goes 260 m up a hill, along a slope of $\theta=17^o$

Therefore vertical elevation of the car = $260sin\theta=260sin17^o=76.0166m$

Now, when you get into the cable car, it's velocity is zero, that is, initial kinetic energy is zero (since K.E. = $\frac{1}{2} mv^2$ ). Similarly as the car reaches the top, it halts and hence final kinetic energy is zero.

Therefore the only possible change in the cable car system is the change in it's gravitational potential energy.

Hence, total change in energy = mgh = $5800\times9.81\times76.0166J=4.325\times10^6J$

where, g = acceleration due to gravity

h = height/vertical elevation

4 0

2 years ago