Ask question

Ask question

All categories

2 years ago

11

At one particular moment, a 15.0 kg toboggan is moving over a horizontal surface of snow at 4.40 m/s. After 7.50 s have elapsed,

the toboggan stops. Use a momentum approach to find the magnitude of the average friction force (in N) acting on the toboggan while it was moving.

1 answer:

MArishka [77]2 years ago

7 0

Answer:

$f = 8.8\,N$

Explanation:

The moment equation of the toboggan is:

$m_{t}\cdot v_{o} - f\cdot \Delta t = 0$

The average friction force is:

$f=\frac{m_{t}\cdot v_{o}}{\Delta t}$

$f = \frac{(15.0\,kg)\cdot (4.40\,\frac{m}{s} )}{7.50\,s}$

$f = 8.8\,N$

You might be interested in

According to Archimedes’ principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this

Andrew [12]

Answer:

Part a)

$\rho = 0.55 g/cm^3$

Part b)

$\rho_L = 1.49 g/cm^3$

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

Explanation:

Part a)

As we know that when cylinder float in the water then weight of the cylinder is counter balanced by the buoyancy force

So here we know

buoyancy force is given as

$F_b = \rho_w V_{sub} g$

$F_b = (1 g/cm^3) (30 - 13.5) Ag$

$F_b = 16.5 Ag$

Now we know that the weight of the cylinder is given as

$W = \rho (30 cm)A g$

now we have

$\rho (30 cm) A g = 16.5 A g$

$\rho = 0.55 g/cm^3$

Part b)

When the same cylinder is floating in other liquid then we will have

$F_b = \rho_L (30 - 18.9 )A g$

so we have

$\rho_L (11.1) Ag = 0.55(30) Ag$

$\rho_L = 1.49 g/cm^3$

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

3 0

2 years ago

Suppose that you'd like to find out if a distant star is moving relative to the earth. The star is much too far away to detect a

rjkz [21]

Answer:

The speed will be "18km/s". A further explanation is given below.

Explanation:

According to the question, the values are:

Wavelength,

$\lambda = 656.46 \ nm$

$\Delta \lambda = 0.04$

$c=3\times 10^8$

As we know,

⇒ $\frac{\Delta \lambda}{\lambda} =\frac{v}{c}$

On substituting the values, we get

⇒ $\frac{656.46}{0.04} =\frac{v}{3\times 10^8}$

⇒ $v=\frac{656.46}{0.04} (3\times 10^8)$

⇒ $=16411.5\times 3\times 10^8$

⇒ $=18280 \ m/s$

or,

⇒ $=18 \ km/s$

8 0

2 years ago

10. Josie and Trey were working on their physics project and both built catapults. Trey's catapult shot a ball

SVEN [57.7K]

Answer:

Josie's ball faster than T

3 0

1 year ago

Using monochromatic light of 410 nm wavelength, a single thin slit forms a diffraction pattern, with the first minimum at an ang

VLD [36.1K]

Answer:

c. 307 nm

Explanation:

angular position of first dark fringe = λ / d , λ is wavelength and d is width of slit .

(40 x π ) / 180 = 410 / d

angular position of second dark fringe = 2 x λ / d , λ is wavelength and d is width of slit .

(60 x π ) / 180 = 2 x λ / d

Dividing these equations

60 / 40 = 2 x λ / 410

λ = 307.5 nm.

5 0

3 years ago

One student did an experiment on the rock cycle.

Nonamiya [84]

First, when the student added the layers of wax over each other, this became a representation of sedimentary rocks.

Then the student folded his/her palm and squeezed the layers of wax. This means that the student applied heat and pressure on the wax (sedimentary rocks)

Referring to the diagram below which represents the rock cycle, we will find that applying heat and pressure on sedimentary rocks would convert these rocks into metamorphic rocks.

Based on the above, the best choice would be:
<span>d. Heat and pressure can change sedimentary rocks into metamorphic rocks.</span>

7 0

2 years ago