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3 years ago

15

A sled with a mass of 32 kg is being pulled across the snow. If the friction force between the sled and the ground is 12 N, what

is the force needed to accelerate the sled at a rate of 1.1 m/s2?

1 answer:

Sergeeva-Olga [200]3 years ago

3 0

Answer:

Fa= 47.2N

Explanation:

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a steel block has a mass of 40g.it is in the form of a cube. each edge is 1.74cm long. calculate the density

Vinil7 [7]

Answer:

d ≈ 7,6 g/cm³

Explanation:

d = m/V = 40g/5,27cm³ ≈ 7,6 g/cm³

V = l³ = (1.74cm)³ ≈ 5,27 cm³

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3 years ago

This is physics class, plzzz help ! 86.72 m to cm

Vaselesa [24]

Answer:

8672

Explanation:

multiply the length value by 100

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3 years ago

22. Explain why it is better to increase the biodiversity of ecosystem.

Marizza181 [45]

Answer: because increasing biodiversity can influence ecosystem functions such as productivity and variety and even the likelihood that a particular species is discovered by a comminity

Explanation:

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3 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0

2 years ago

An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance be

Inessa05 [86]

Explanation:

The given data is as follows.

Inner wheel Radius = 20 m,

Distance between left and right wheel = 1.5m,

Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

Angular velocity of automobile = w = $\frac{V}{R}$

where, V = linear velocity of automobile m/min,

R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

u = linear velocity of inner wheel

and, u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

w =

= $\frac{u}{(R - d)}$

u = $V \times \frac{(R - d)}{R}$

= $V \times (1 - \frac{d}{R})$

If radius of wheel is r it will cover distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus, u = $2\pi rn$ = $V \times (1 - \frac{d}{R})$

Now, rotation per minute of inner wheel is calculated as follows.

n = $\frac{V}{2 \pi r \times (1 - \frac{d}{R})}$

= $\frac{V}{2 \pi r \times (1 - \frac{0.75}{20})}$ (since 2d = 1.5m given, d = 0.75m),

= $\frac{V}{r} \times 0.1532$

So, rotation per minute of outer wheel; n' =

= $\frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}$

= $\frac{V}{r} \times 0.1651$

5 0

3 years ago