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2 years ago

Suppose a 0.04-kg car traveling at 2.00 m/s can barely break an egg. What is the min

Physics

1 answer:

Paha777 [63]2 years ago

4 0

Answer: The correct answer is option (A).

Explanation:

Momentum of the car with 0.04 kg mass , which travelling with velocity of 2.00 m/s

$P_1=mass\times velocity=m_1\times v_1=0.04 kg\times 2.00 m/s=8.00 kg m/s$

Then the maximum speed of the another car in order to not to break the eggs will be same as first car:

$P_1=P_2$

$0.08 kg m/s=m_2\times v_2=0.08 kg\times v_2$

$v_2=1 m/s$

Speed slightly more than 1 m/s will increase the momentum of second car and the eggs will break. So, from the given options the minimum speed need by the second car will be 1.42m/s.

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Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.340 m and carries a current

sladkih [1.3K]

Answer:

The y-value of the line in the xy-plane where the total magnetic field is zero $U = 0.1355 \ m$

Explanation:

From the question we are told that

The distance of wire one from two along the y-axis is y = 0.340 m

The current on the first wire is $I_1 = (27.5i) A$

The force per unit length on each wire is $Z = 295 \mu N/m = 295*10^{-6} N/m$

Generally the force per unit length is mathematically represented as

$Z = \frac{F}{l} = \frac{\mu_o I_1I_2}{2\pi y}$

=> $\frac{\mu_o I_1I_2}{2\pi y} = 295$

Where $\mu_o$ is the permeability of free space with a constant value of $\mu_o = 4\pi *10^{-7} \ N/A2$

substituting values

$\frac{ 4\pi *10^{-7} 27.5 * I_2}{2\pi * 0.340} = 295 *10^{-6}$

=> $I_2 = 18.23 \ A$

Let U denote the line in the xy-plane where the total magnetic field is zero

So the force per unit length of wire 2 from line U is equal to the force per unit length of wire 1 from line (y - U)

$\frac{\mu_o I_2 }{2 \pi U} = \frac{\mu_o I_1 }{2 \pi(y - U) }$

substituting values

$\frac{ 18.23 }{ U} = \frac{ 27.5 }{(0.34 - U) }$

$6.198 -18.23U = 27.5U$

$6.198=45.73U$

$U = 0.1355 \ m$

5 0

2 years ago

In the demolition of an old building, a 1,300 kg wrecking ball hits the building at 1.07 m/s2. Calculate the amount of force at

Y_Kistochka [10]

Answer: F = 1391 N

Explanation:

The information given to you are:

Mass M = 1300 kg

Acceleration a = 1.07 m/s^2

The magnitude of the force striking the building will be

F = ma

Where

F = force

Substitute mass M and acceleration a into the formula

F = 1300 × 1.07

F = 1391 N

Therefore, the wrecking ball strikes the building with a force of 1391 N

3 0

2 years ago

A farsighted girl has a near point at 2.0 m but has forgotten her glasses at home. The girl borrows eyeglasses that have a power

lisov135 [29]

Find the pictures in the attachment

6 0

2 years ago

The banking angle in a turn on the Olympic bobsled track is not constant, but increases upward from the horizontal. Coming aroun

zlopas [31]

Answer:

the curve inclination is increased so that a weight component helps keep the car on track

Explanation:

In the sledging competition these devices go at quite high speeds over 100 km/h, so when reaching the curves the friction force is not enough to keep the car on the track. For this reason, the curve inclination is increased so that a weight component helps keep the car on track.

In general we can solve Newton's second law for this case, with the condition of no friction, it is found that

V² = r g tan θ

Where V is the maximum velocity, r is the radius of the curve a, θ is the angle of the inclination

8 0

2 years ago

Dane is holding an 8 kilogram box 2 metres above the ground. How much energy is in the box's gravitational potential energy stor

9966 [12]

156.8 Joules of energy is in the box's gravitational potential energy store

Explanation:

Given:

Mass of the box Dane is holding = 8 Kilograms

Height at which Dane is holding the box above the ground= 2 metres

To Find:

Gravitational potential energy in the box=?

Solution:

gravitational potential energy is the work done per mass on a object to move that object from one fixed location to to another location against gravity.Its unit is joules or J

Thus Gravitational potential energy is represented as,

$PE_g=mgh$