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Veseljchak [2.6K]
3 years ago
10

Which of the following is true about the easy walk harness

Physics
2 answers:
Svetllana [295]3 years ago
6 0
<span>the easy walk harness should be looser on the dog than other harnesses</span>
tiny-mole [99]3 years ago
3 0

Answer:

The answer is option A

the leash always attaches to the easy walk harness in the front.

Explanation:

Attaching the leash to both the chest ring of the harness AND the leash ring of a regular collar helps to keep the harness in place and also prevents escape. See the diagram below


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A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. if the ball hits the ground 4.0 second
Viktor [21]

Answer:

78.4 m

Explanation:

To obtain the height of the cliff;

We can use the Relation to obtain the final velocity, v

v = u + at

a = acceleration due to gravity = 9.8m/s²

v = 0 + (9.8*4)

v = 0 + 39.2

v = 39.2 m/s

To obtain the Height, S

v² = u² + 2aS

39.2^2 = 0 + 2(9.8)S

39.2^2 = 0 + 19.6S

1536.64 = 19.6S

S = 1536.64 / 19.6

S = 78.4 m

8 0
2 years ago
How much current is in a circuit that includes a 9.0 volt battery and a bulb woth a resistance of 4.0 ohms
Vitek1552 [10]

According to Ohm's law for a portion of the circuit we have:

U=RI=>I=U/R=9/4=2.25 A

7 0
3 years ago
Read 2 more answers
A particle of charge 2.3 ✕ 10−8 C experiences an upward force of magnitude 4.6 ✕ 10−6 N when it is placed in a particular point
Marysya12 [62]
<h2>Answer:</h2>

(a) +2 x 10² N/C (upwards)

(b) -2.2μN or -2.2 x 10⁻⁶N (downwards)

<h2>Explanation:</h2>

The force (F) acting on a particle of charge (Q) at a particular point is related to its electric field (E) by the following;

F = Q x E   ----------------------(i)

This means that the force acting on the charged particle is the product of its charge and the electric field at that point.

<em>(a) Given</em>;

Q = charge of the particle = 2.3 x 10⁻⁸ C

F = force acting on the particle = 4.6 x 10⁻⁶N

<em>Substitute these values into equation (i) as follows;</em>

=> 4.6 x 10⁻⁶  = 2.3 x 10⁻⁸ x E

=> E = 4.6 x 10⁻⁶ ÷ 2.3 x 10⁻⁸

=> E =  2 x 10² N/C

Since the value is positive, the electric field is directed upwards.

Therefore, the electric field at that point is +2 x 10² N/C

<em>(b) If a charge of q = -1.1 x 10⁻⁸ is placed there, the force (F) acting is calculated as follows;</em>

<em>Substitute Q = q into equation (i) as follows;</em>

F = q x E

<em>Substitute the value of q and E = 2 x 10² N/C into the equation above as follows;</em>

F = -1.1 x 10⁻⁸ x 2 x 10²

F = -2.2 x 10⁻⁶ N

F = -2.2μN

Since the value of the force, F, is negative, it means it is directed downwards.

Therefore the force on the charge is -2.2μN

3 0
3 years ago
In a double-slit experiment, the third-order maximum for light of wavelength 510 nm is located 17 mm from the central bright spo
Marysya12 [62]

Answer:

14.9 mm

Explanation:

We know dsinθ = mλ where d = separating of slit, m = order of maximum = 3 and λ = wavelength = 510nm = 510 × 10⁻⁹ m

Also tanθ = L/D where L = distance of m order fringe from central bright spot = 17 mm = 0.017 m and D = distance of screen from slit = 1.6 m

So, sinθ = mλ/d

Since θ is small, sinθ ≅ tanθ

So,

mλ/d = L/D

d = mλD/L

Substituting the values of the variables into the equation, we have

d = 3 × 510 × 10⁻⁹ m × 1.6 m/0.017 m

d = 2448 × 10⁻⁹ m²/0.017 m

d = 144000 × 10⁻⁹ m

d = 1.44 × 10⁻⁴ m

d = 0.144 × 10⁻³ m

d = 0.144 mm

Now, for the second-order maximum, m' of the 670 nm wavelength of light,

m'λ'/d = L'/D where m' = order of maximum = 2, λ' = wavelength of light = 670 nm = 670 × 10⁻⁹ m, d = slit separation = 0.144 mm = 0.144 × 10⁻³ m, L' = distance of second order maximum from central bright spot and D = distance of screen from slit = 1.6m

So, L' = m'λ'D/d

So, substituting the values of the variables into the equation, we have

L' = 2 × 670 × 10⁻⁹ m × 1.6 m/0.144 × 10⁻³ m

L' = 2144  × 10⁻⁹ m²/0.144 × 10⁻³ m

L' = 14888.89 × 10⁻⁶ m

L' = 0.01488 m

L' ≅ 0.0149 m

L' = 14.9 mm

4 0
3 years ago
A book rest on a table which it's face having a sides 30cm by 25cm . if it exerts apressure of 200pa then determine the mass of
Simora [160]

Answer:

Approximately 1.5\; \rm kg. (Assuming that this table is level, and that the gravitational field strength is g = 9.8\; \rm N \cdot kg^{-1}.

Explanation:

Convert the dimensions of this book to standard units:

\displaystyle 30\; \rm cm = 30\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm}  = 0.30\; \rm m.

\displaystyle 25\; \rm cm = 25\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm}  = 0.25\; \rm m.

Calculate the surface area of this book:

0.30\; \rm m  \times 0.25\; \rm m = 0.075\; \rm m^{2}.

Pressure is the ratio between normal force and the area over which this force is applied.

\displaystyle \text{Pressure} = \frac{\text{normal Force}}{\text{contact Area}}.

Equivalently:

(\text{normal Force}) = \text{Pressure} \cdot (\text{contact Area}).

In this question, \text{Pressure} = 200\; \rm Pa = 200\; \rm N \cdot m^{-2}.

It was found that (\text{contact Area}) = 0.075\; \rm m^{2} (assuming that the entire side of this book is in contact with the table.

Hence:

\begin{aligned}& (\text{normal Force}) \\ &= \text{Pressure} \cdot (\text{contact Area}) \\ &= 200\; \rm N \cdot m^{-2} \times 0.075\; \rm m^{2} \\ &= 15\; \rm N \end{aligned}.

If that the table is level, this normal force would be equal to the weight of this book:

\text{weight} = 15\; \rm N.

Assuming that the gravitational field strength is g = 9.8\; \rm N \cdot kg^{-1}. The mass of this book would be:

\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{15\; \rm N}{9.8\; \rm N \cdot kg^{-1}}\approx 1.5\; \rm kg\end{aligned}.

4 0
3 years ago
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