All categories

2 years ago

What is the maximum negative displacement a dog could have if it started its motion at +3 m?

Physics

1 answer:

raketka [301]2 years ago

7 0

Answer:

- 3 meter

Explanation:

A dog has started motion from +3 meter. ...(Given)

∴ maximum positive distance = + 3 meter

Magnitude of distance = 3

Maximum negative distance = (-) (magnitude of distance)

Maximum negative distance = (-) (3)

Maximum negative distance = -3 meter

Hence, maximum negative distance is -3 meter.

You might be interested in

A worker wants to turn over a uniform 1110-N rectangular crate by pulling at 53.0 ∘ on one of its vertical sides (the figure (Fi

tekilochka [14]

This problem has three questions I believe:

> How hard does the floor push on the crate?

We have to find the net vertical (normal) Fn force which results from Fp and Fg.
We know that the normal component of Fg is just Fg, which is equal to as 1110N.
From the geometry, the normal component of Fp can be calculated:
Fpn = Fp * cos(θp)
= 1016.31 N * cos(53)
= 611.63 N

The total normal force Fn then is:
Fn = Fg + Fpn
= 1110 + 611.63
= 1721.63 N

> Find the friction force on the crate

We have to look for the net horizontal force Fh which results from Fp and Fg. Since Fg is a normal force entirely, so we can say that the horizontal component is zero:

Fh = Fph + Fgh
= (Fp * sin(θp)) + 0
= 1016.31 N * sin(53)
= 811.66 N

> What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

We just need to compute the ratio Fh to Fn to get the minimum μs.

μs = Fh / Fn

= 811.66 N / 1721.63 N

= 0.47

8 0

2 years ago

The inventor of the photographic process in which a photograph produced without a negative by exposing objects to light on light

Nikolay [14]

A. Man Ray --------------------

7 0

1 year ago

Read 2 more answers

A 12 kg box sliding on a horizontal floor has an initial speed of 4.0 m/s. The coefficient of friction bctwecn thc box and the f

Hitman42 [59]

Answer:

(D) 96 kg-m/s

Explanation:

Let's start off by first calculating the normal force between the box and the floor.

This will be:

Normal Force = 12 * 9.81 = 117.72 N

We can now use the friction equation to find the frictional force on the box when it is moving:

Frictional force = Coefficient of friction * Normal Force

Frictional force = 0.4 * 117.72 = 47.09 N

Finally, since we have the force on the box, we can find the acceleration:

F = Mass * Acceleration

47.09 = 12 * Acceleration

Acceleration = 3.92 m/s^2

Final speed after 2 seconds:

$V=U+a*t$

$V = 4 +(-3.92)*(2)$

V= -3.84 m/s

Since we know the initial and final speeds, we can calculate the change in momentum:

Change in momentum = Final Momentum - Initial Momentum

Change in momentum = $3.84*12-(-4)*12$

Change in momentum = 94.08 kg*m/s

Thus we can see that option (D) is the closest answer.

6 0

2 years ago

Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d

Step2247 [10]

Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that $F = k \frac{Q_1Q_2}{d^2}$ so if we double the charge on $Q_1$ and double the distance to $(2d)$ we plug these into the equation to find

 $F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}$ 

So we see the new force is exactly 1/2 of the old force so your answer should be $\frac{1}{2}F$ if I can remember my physics correctly.

9 0

2 years ago

Read 2 more answers

A microwave oven operates with sinusoidal microwaves at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the

Degger [83]

Answer:

F = 2 × 10⁻³ N

Explanation:

Given:

frequency, f = 2400 MHz

Height, h = 25cm = 0.25 m

Area of the base, A = 30 cm x 30 cm = 900 cm² = 0.09 m²

Energy of the microwave, E = 0.50 mJ = 0.5 x 10⁻³ J

Now, the time taken by the wave from top to the base, t = h/c

here, c is the speed of the light

thus,

t = 0.25/(3 x 10⁸) = 8.33 x 10⁻¹⁰ s

The radiation pressure $P_r$ = Intensity/c

now, the intensity is given as:

I = Power/ area

also,

Power = Energy/ time = 0.5 x 10⁻³ J/8.33 x 10⁻¹⁰ s = 600000 W

thus,

I = 600000 W/ 0.09 m² = 6666666.6 W/m²

substituting the value in the formula for pressure due to radiation, we have

$P_r$ = 6666666.6 W/m²/(3 x 10⁸)

also

pressure = Force/ area

thus,

Force/ area = 6666666.6 W/m²/(3 x 10⁸)

Force (F) = (6666666.6 W/m² × 0.09 m²)/(3 x 10⁸)

F = 2 × 10⁻³ N

6 0

2 years ago

Read 2 more answers