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2 years ago

A child drags a 0.398 kg toy dog

Physics

1 answer:

Leviafan [203]2 years ago

8 0

Answer:

2.10 N

Explanation:

Draw a free body diagram of the toy. There are four forces:

Weight force mg pulling down,

Normal force N pushing up,

Friction force F pushing left,

Applied force 4.63 N pulling at an angle of 63.0°.

Sum of the forces in the x direction:

∑F = ma

4.63 N cos 63.0° − F = 0

F = 2.10 N

You might be interested in

500mL of He at 98 kPa expands to 750 mL. Find P2

GarryVolchara [31]

The value of P₂ is 65.33 kPa

<u>Explanation:</u>

Given:

Volume, V₁ = 500mL

Pressure, P₁ = 98 kPa

Volume, V₂ = 750 mL

Pressure, P₂ = ?

According to Boyle's law:

P₁V₁ = P₂V₂

On substituting the value we get:

$98 X 500 = 750 X P_2\\\\P_2 = \frac{98 X 500}{750} \\\\P_2 = 65.33 kPa$

The value of P₂ is 65.33 kPa

3 0

2 years ago

(a) Triply charged uranium-235 and uranium-238 ions are being separated in a mass spectrometer. (The much rarer uranium-235 is u

stiv31 [10]

Answer:

(a) 2.5 cm

(b) Yes

Solution:

As per the question:

Mass of Uranium-235 ion, m = $3.95\times 10^{- 25}\ kg$

Mass of Uranium- 238, m' = $3.90\times 10^{- 25}\ kg$

Velocity, v = $3.00\times 10^{5}\ m/s$

Magnetic field, B = 0.250 T

q = 3e

Now,

To calculate the path separation while traversing a semi-circle:

$\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})$

The radius of the ion in a magnetic field is given by:

R = $\frac{mv}{qB}$

$\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})$

$\Delta x = 2(\frac{mv}{qB} - \frac{m'v}{qB})$

$\Delta x = 2(\frac{m - m'}{qB}v)$

Now,

By putting suitable values in the above eqn:

$\Delta x = 2(\frac{3.95\times 10^{- 25} - 3.90\times 10^{- 25}}{3\times 1.6\times 10^{- 19}\times 0.250}\times 3.00\times 10^{5}) = 2.5\ cm$

$\Delta x = 1.25\ cm$

(b) Since the order of the distance is in cm, thus clearly this distance is sufficiently large enough in practical for the separation of the two uranium isotopes.

3 0

2 years ago

The voltage across the primary winding is 6000 V. If the primary winding has 50 coils and the secondary winding has 20 coils, wh

sesenic [268]

Answer: 2400V

Explanation:

This is a question on transformer. A transformer is a device used to step up or step down current and voltage. It consists of the primary windings and secondary windings. Each windings possesses both coil and amount of voltage across each coil.

The transformer formula is

Vs/Vp = Ns/Np

Where;

Vs is the voltage across the secondary coil

Vp is the voltage across the primary coil

Ns is the number of turn in the primary coil

Np is the number of turns in the primary coil.

According to the question,

Vs is unknown

Vp is 6000V

Ns is 20turns

Np is 50turns

Substituting the values in the formula given

Vs/6000 = 20/50

Vs = 6000×20/50

Vs = 2400V

Therefore the secondary voltage is 2400V

4 0

2 years ago

Read 2 more answers

Sheila (m=56.8 kg) is in her saucer sled moving at 12.6 m/s at the bottom of the sledding hill near Bluebird Lake. She approache

FromTheMoon [43]

Answer:

y = 54.9 m

Explanation:

For this exercise we can use the relationship between the work of the friction force and mechanical energy.

Let's look for work

W = -fr d

The negative sign is because Lafourcade rubs always opposes the movement

On the inclined part, of Newton's second law

Y Axis

N - W cos θ = 0

The equation for the force of friction is

fr = μ N

fr = μ mg cos θ

We replace at work

W = - μ m g cos θ d

Mechanical energy in the lower part of the embankment

Em₀ = K = ½ m v²

The mechanical energy in the highest part, where it stopped

$Em_{f}$ = U = m g y

W = ΔEm = $Em_{f}$ - Em₀

- μ m g d cos θ = m g y - ½ m v²

Distance d and height (y) are related by trigonometry

sin θ = y / d

y = d sin θ

- μ m g d cos θ = m g d sin θ - ½ m v²

We calculate the distance traveled

d (g syn θ + μ g cos θ) = ½ v²

d = v²/2 g (sintea + myy cos tee)

d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)

d = 1555.85 /7.8145

d = 199.1 m

Let's use trigonometry to find the height

sin 16 = y / d

y = d sin 16

y = 199.1 sin 16

y = 54.9 m

8 0

2 years ago

You and your friend throw balloons filled with water from the roof of a several story apartment house. You simply drop a balloon

Aleks [24]

Answer:

Height = 53.361 m

Explanation:

There are two balloons being thrown down, one with initial speed (u1) = 0 and the other with initial speed (u2) = 43.12

From the given information we make the following summary

$u_{1}$ = 0m/s

$t_{1}$ = t

$u_{2}$ = 43.12m/s

$t_{2}$ = (t-2.2)s

The distance by the first balloon is

$D = u_{1} t_{1} + \frac{1}{2} at_{1}^2$

where

a = 9.8m/s2

Inputting the values

$D = (0)t + \frac{1}{2} (9.8)t^2\\ D = 4.9t^2$

The distance traveled by the second balloon

$D = u_{2} t_{2} + \frac{1}{2} at_{2}^2$

Inputting the values

$D = (43.12)(t-2.2) + \frac{1}{2} (9.8)(t-2.2)^2$

simplifying

$D = 4.9t^2 + 21.56t -71.148$

Substituting D of the first balloon into the D of the second balloon and solving

$4.9t^2 = 4.9t^2 + 21.56t -71.148 \\ 21.56t = 71.148\\ t = 3.3s$

Now we know the value of t. We input this into the equation of the first balloon the to get height of the apartment

$D = 4.9(3.3)^2\\ D = 53.361 m$

7 0

2 years ago