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2 years ago

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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. Find the acceleration of the sleds.

1 answer:

KonstantinChe [14]2 years ago

6 0

Answer:

Acceleration, $a=1.2\ m/s^2$

Explanation:

Given that,

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

Force exerted by the doges on the rope attached to the front sled, F = 240 N

To find,

The acceleration of the sleds.

Solution,

Let a is the acceleration of the sleds. The product of mass and acceleration is called force. Its expression is given by :

F = ma

$a=\dfrac{F}{m}$

$a=\dfrac{240\ N}{2\times 100\ kg}$ (m = 2m)

$a=1.2\ m/s^2$

So, the acceleration of the sleds is $1.2\ m/s^2$ .

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The wavelength of green light is 550 nm.

SIZIF [17.4K]

Answer:

(a) momentum of photon is 1.205 x 10⁻²⁷ kgm/s

velocity of electron is 1323.88 m/s

momentum of the electron is 1.205 x 10⁻²⁷ kgm/s

(b) momentum of photon is 1.506 x 10⁻²⁷ kgm/s

velocity of electron is 1654.85 m/s

momentum of the electron is 1.506 x 10⁻²⁷ kgm/s

(c) The momentum of the photon is equal to the momentum of the electron

Explanation:

(a)

wavelength of green light, λ = 550 nm

momentum of photon is given by;

$p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{550*10^{-9}}\\\\p = 1.205 *10^{-27} \ kg.m/s$

velocity of electron is given by;

$P = \frac{h}{\lambda} \\\\mv = \frac{h}{\lambda}\\\\v = \frac{h}{m \lambda}\\\\v = \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(550*10^{-9})}\\\\v = 1323.88 \ m/s$

momentum of the electron is given by;

p = mv

p = (9.1 x 10⁻³¹) (1323.88)

p = 1.205 x 10⁻²⁷ kgm/s

(b)

wavelength of red light, λ = 440 nm

momentum of photon is given by;

$p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{440*10^{-9}}\\\\p = 1.506 *10^{-27} \ kg.m/s$

velocity of electron is given by;

$v = \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(440*10^{-9})}\\\\v = 1654.85 \ m/s$

momentum of the electron is given by;

p = mv

p = (9.1 x 10⁻³¹) (1654.85)

p = 1.506 x 10⁻²⁷ kgm/s

(c) The momentum of the photon is equal to the momentum of the electron.

7 0

2 years ago

Buffalo, New York, experienced a snowstorm November 13–21, 2014. Residents refer to the event as “Snowvember.” What was the like

scZoUnD [109]

I know you're probably done with this by now, but the answer is *Lake-Effect Snow*

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2 years ago

Read 2 more answers

When boiling water, a hot plate takes an average of 8 minutes and 55 seconds to boil 100 milliliters of water. Assume the temper

alexandr1967 [171]

Answer:

90.9 seconds

Explanation:

m = Mass of liquid = Volume×Density

c = Specific heat

$\Delta T$ = Change in temperature

t = Time taken

Room temperature = 75 °F

Converting to Celsius

$(75-32)\times \frac{5}{9}=23.889\ ^{\circ}C$

Heat required to raise the temperature of water

$Q=mc\Delta T\\\Rightarrow Q=100\times 10^{-6}\times 1000\times 4186\times (100-23.889)\\\Rightarrow Q=31860.0646\ J$

Power

$P=\frac{Q}{t}\\\Rightarrow P=\frac{31860.0646}{8\times 60+55}\\\Rightarrow P=59.55152\ W$

Efficiency of the plate

$\frac{59.5512}{283}\times 100=21.04282\%$

Heat required to raise the temperature of water

$Q=mc\Delta T\\\Rightarrow Q=100\times 10^{-6}\times 784\times 2150\times (56-23.889)\\\Rightarrow Q=5412.63016\ J$

$P=\frac{Q}{t}\\\Rightarrow t=\frac{Q}{P}\\\Rightarrow t=\frac{5412.63016}{0.2104282\times 283}\\\Rightarrow t=90.9\ s$

Time taken to heat the aceton is 90.9 seconds

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2 years ago

A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force

VikaD [51]

Answer:

a) the values of the angle α is 45.5°

b) the required magnitude of the vertical force, F is 41 lb

Explanation:

Applying the free equilibrium equation along x-direction

from the diagram

we say

∑Fₓ = 0

Pcosα - 425cos30° = 0

525cosα - 368.06 = 0

cosα = 368.06/525

cosα = 0.701

α = cos⁻¹ (0.701)

α = 45.5°

Also Applying the force equation of motion along y-direction

∑Fₓ = ma

Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)

525sin45.5° + F + 212.5 - 600 = 27.95

374.46 + F + 212.5 - 600 = 27.95

F - 13.04 = 27.95

F = 27.95 + 13.04

F = 40.99 ≈ 41 lb

8 0

2 years ago

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direc

Alborosie

The given question is incomplete. The complete question is as follows.

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction.

A) If the net force acting on the particle is $6.21 \times 10^{-3}$ N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the x-y plane.

Enter your answers numerically separated by commas.

Explanation:

The given data is as follows.

Q = $6.50 \times 10^{-6} C$

E = 1300 N/C in the +x direction

B = 1.02 T in the +z direction

and, $F_{net} = 6.25 \times 10^{-3} N$ in the +x direction

Also, $F_{net} = F_{E} - F_{b}$

= qE - qvB

Now, we will calculate the value of v as follows.

v = $(\frac{1}{B}) \times (E - \frac{F_{net}}{q})$

= $(\frac{1}{1.02 T}) \times (1300 - \frac{6.25 \times 10^{-3}}{6.50 \times 10^{-6}})$

v = 458.507 m/s

Using the value for velocity, we need to know which direction it's going.

You know +x direction for E, +z direction for B and +x for $F_{net}$ .

Using the right hand rule where:

your right thumb goes toward the $F_{net}$ , then your index finger points to B (z direction) Then curl your middle, ring, and pink 90 angle. This shows where v is going which is -y direction.

Thus, we can conclude that $v_{x}, v_{y}, v_{z}$ = 0, -(458.507), 0.

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2 years ago