All categories

3 years ago

This question is typical on some driver’s license exams: A car moving at 49 km/h skids

Physics

1 answer:

bearhunter [10]3 years ago

3 0

42.5 m far the car would skid with locked brakes at speed of 122.5 km/h.

Explanation:

As per the given question, friction force is constant

A car moving at 49 km/h skids 17 m

$\text { So speed } S_{1}=49 \mathrm{km} / \mathrm{h}$ and

$\text { Distance } \mathrm{D}_{1}=17 \mathrm{m}$

The car skid with locked brakes at 122.5 km/h "x" distance.

$\text { So speed } \mathrm{S}_{2}=122.5 \mathrm{km} / \mathrm{h}$

Distance $\mathrm{D}_{2}=\text { unknown }(\mathrm{x})$

We know that ratio of distance and speed is

$\frac{d_{1}}{d_{2}}=\frac{s_{1}}{s_{2}}$

$\frac{17}{d_{2}}=\frac{49}{122.5}$

Cross multiply

$d_{2}=\frac{17 \times 122.5}{49}$

$\mathrm{d}_{2}=\frac{2082.5}{49}$

$\mathrm{d}_{2}=42.5 \mathrm{m}$

The distance skid by the car with locked brakes at speed of 122.5 km/h is 42.5 m.

You might be interested in

The direction of rotation of Venus is _________ to the direction of all other planets.

MakcuM [25]

A - opposite

I have to write 20 characters so....jghghgyf

7 0

3 years ago

Two long, parallel wires are separated by 2.0 m. Each wire has a 28-A current, but the currents are in opposite directions. Part

BlackZzzverrR [31]

Given Information:

Current = I = 28 A

distance between wires = r = 2.0 m

Required Information:

Magnetic field = B = ?

Answer:

B = 12x10⁻⁶ T

Step-by-step explanation:

Biot-Savart Law is given by

B = μ₀I/2πr

Where μ₀ is the permeability of free space, I is the current flowing through the wire and B is the magnitude of the magnetic field produced.

We are asked to find the magnetic field midway between the wires so r/2 = 1

B = 4πx10⁻⁷*28/2π*1

B = 6x10⁻⁶ T

since the same amount of current flows in both wires therefore, equal amount of magnetic field will be produced in both wires

B = 2*6x10⁻⁶ T

B = 12x10⁻⁶ T

Therefore, the net magnetic field midway between the two wires is 12x10⁻⁶ T.

3 0

3 years ago

A 10 m long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 30 m/s when a 68

timama [110]

Answer:

v = 30.39 m/s

Explanation:

given,

mass of glider,M = 680 Kg

mass of the skydiver, m = 68 Kg

horizontal velocity,V = 30 m/s

when skydiver releases, velocity,v = 30 m/s

velocity if the glider,v' = ?

use the conservation of momentum

M V = m' v' + m v

m' = 680-68 = 612 Kg

680 x 30 = 612 x v + 60 x 30

612 v = 18600

v = 30.39 m/s

since the skydiver's speed will be the same as before release

The glider should continue to travel at 30.39 m/s since there are no external forces acting on it

5 0

3 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

rjkz [21]

Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

$F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$