2) A bully is pushing a boy against a locker as shown with a force of

Time taken can be calculated when distance and the speeds are given. Here speeds of Keith and Sue are given. So, we have to find the relative speeds in order to calculate the time taken.

When two objects travel in same direction the relative speed will be the difference between speeds. Similarly when two objects travel in opposite direction, the relative speed will be the sum of given speeds.

Given:

Speed of Sue from Burkey is 6 km/hr and speed of Keith from Baron is 3 km/hr.

The distance between Burkey and Baron is 300 m.

From the formula, $d=s \times t$

where d is distance,s is speed and t is time

It can be derived that, $t=\frac{d}{s}$

s = sum of given speeds = 3 km/hr + 6 km/hr = 9 km/hr

d = 300 m = 0.3 km $\text {Time }=\frac{0.3 \mathrm{km}}{9 \mathrm{km} / \mathrm{hr}}=\frac{1}{30} h r=2 \text { minutes }$

3 0

4 years ago

A proton moves with a velocity of v with arrow = (3î − 5ĵ + k) m/s in a region in which the magnetic field is B with arrow = (î

Allisa [31]

Answer:

The magnitude of the magnetic force this particle experiences is $2.6\times10^{-9}\ N$ .

Explanation:

Given that,

Velocity v= (3i-5j+k) m/s

Magnetic field B=(i+2j-k) T

We need to calculate the value $\vec{v}\times\vec{B}$

$(\vec{v}\times\vec{B})=3i+4j+11k$

We need to calculate the magnitude of the magnetic force this particle experiences

Using formula of magnetic force

$\vec{F}=q(\vec{v}\times\vec{B})$

Put the value into the formula

$\vec{F}=1.6\times10^{-19}\times(3i+4j+11k)$

$\vec{F}=(4.8i+6.4j+1.76k)\times10^{-19}$

$|F|=2.6\times10^{-9}\ N$

Hence, The magnitude of the magnetic force this particle experiences is $2.6\times10^{-9}\ N$ .

7 0

3 years ago

A 2.4 kg box has an initial velocity of 3.6 m/s upward along a plane inclined at 27◦ to the horizontal. The coefficient of kinet

Vika [28.1K]

Answer:

d= 1.18 m

Explanation:

In abscense of friction, total mechanical energy must be constant, i.e.,

ΔK + ΔU = 0

As we are told that there exists a kinetic friction between the box and the plane, we need to take into account the work done by the friction force in the equation, as follows:

ΔK + ΔU = Wnc (1)

If we take as our zero gravitational potential energy reference, the height at which the box is sent upward, we can write the following equations for the different terms in (1):

ΔK = Kf- K₀ = 0 - 1/2*m*v₀² = -1/2*2.4kg* (3.6)²(m/s)² = -15.6 J

ΔU = Uf - U₀ = m*g*h = *m*g*d*sin θ = 2.4 kg*9.81 m/s²*d*0.454 = 10.7*d J

Wnc = Ff. d* cos (180º) = μk*N*d*cos(180º) (2)

The friction force always opposes to the displacement, so the angle between force and displacement is 180º.

The normal force, as is always perpendicular to the surface, takes the value needed to equilibrate the component of the weight perpendicular to the incline, as follows:

N = m*g*cosθ = 2.4 kg*9.81 m/s²*cos 27º = 21 N

Replacing in (2):

Wnc = 0.12*21*cos (180º) = -2.52*d J

Replacing in (1):

-15.6 J + 10.7*d J = -2.52*d J

Solving for d:

d = 1.18 m

7 0

3 years ago