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melisa1 [442]
2 years ago
6

You are riding n a bus moving slowly through heavy traffic at 2.0 m/s. You hurry to the front of the bus at 4.0 m/s relative to

the bus. What is your speed relative to the street?
Physics
1 answer:
kati45 [8]2 years ago
7 0

Answer:

6 m/s

Explanation:

if the bus and you are moving in the same direction, then you add your speeds together

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A _____ has to do with the direction of a force​
Elenna [48]

Answer:

The "solid force"? ... The direction of the force always seems to be coming out of the solid surface. A direction which is perpendicular to the plane of a surface is said to be normal. The force that a solid surface exerts on anything in the normal direction is called the normal force.

Explanation:

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Time Intervals in Ice Ages
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It started off with 68% less than it did at the peak, and later created a void and melted the remainder of the ice at about 92%
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3 years ago
A wave has frequency of 50 Hz and a wavelength of 10 m. What is the speed?
Sunny_sXe [5.5K]

Answer:

v=500 m/s

<h3>Solution:</h3>

v=(10m)(50/s)

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2 years ago
The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23
Burka [1]

Answer:

a) v(2) = 3.9\,\frac{m}{s}, b) v(4) = -15.7\,\frac{m}{s}

Explanation:

a) The equation for vertical velocity is obtained by deriving the function with respect to time:

v(t) = 23.5 -9.8\cdot t

The velocities at given instants are, respectivelly:

v(2) = 3.9\,\frac{m}{s}

v(4) = -15.7\,\frac{m}{s}

8 0
3 years ago
In a second experiment, you decide to connect a string which has length L from a pivot to the side of block A (which has width d
Salsk061 [2.6K]

Answer:

The answer is in the explanation

Explanation:

A)

i) The blocks will come to rest when all their initial kinetic energy is dissipated by the friction force acting on them. Since block A has higher initial kinetic energy, on account of having larger mass, therefore one can argue that block A will go farther befoe coming to rest.

ii) The force on friction acting on the blocks is proportional to their mass, since mass of block B is less than block A, the force of friction acting on block B is also less. Hence, one might argue that block B will go farther along the table before coming to rest.

B) The equation of motion for block A is

m_{A}\frac{\mathrm{d} v}{\mathrm{d} t} = -m_{A}g\nu_{s}\Rightarrow \frac{\mathrm{d} v}{\mathrm{d} t} = -\nu_{s}g \quad (1)

Here, \nu_{s} is the coefficient of friction between the block and the surface of the table. Equation (1) can be easily integrated to get

v(t) = C-\nu_{s}gt \quad (2)

Here, C is the constant of integration, which can be determined by using the initial condition

v(t=0) = v_{0}\Rightarrow C = v_{0} \quad (3)

Hence

v(t) = v_{0} - \nu_{s}gt \quad (4)

Block A will stop when its velocity will become zero,i.e

0 = v_{0}-\nu_{s}gT\Rightarrow T = \frac{v_{0}}{\nu_{s}g} \quad (5)

Going back to equation (4), we can write it as

\frac{\mathrm{d} x}{\mathrm{d} t} = v_{0}-\nu_{s}gt\Rightarrow x(t) = v_{0}t-\nu_{s}g\frac{t^{2}}{2}+D \quad (6)

Here, x(t) is the distance travelled by the block and D is again a constant of integration which can be determined by imposing the initial condition

x(t=0) = 0\Rightarrow D = 0 \quad (7)

The distance travelled by block A before stopping is

x(t=T) = v_{0}T-\nu_{s}g\frac{T^{2}}{2} = v_{0}\frac{v_{0}}{\nu_{s}g}-\nu_{s}g\frac{v_{0}^{2}}{2\nu_{s}^{2}g^{2}} = \frac{v_{0}^{2}}{2\nu_{s}g} \quad (8)

C) We can see that the expression for the distance travelled for block A is independent of its mass, therefore if we do the calculation for block B we will get the same result. Hence the reasoning for Student A and Student B are both correct, the effect of having larger initial energy due to larger mass is cancelled out by the effect of larger frictional force due to larger mass.

D)

i) The block A is moving in a circle of radius L+\frac{d}{2} , centered at the pivot, this is the distance of pivot from the center of mass of the block (assuming the block has uniform mass density). Because of circular motion there must be a centripetal force acting on the block in the radial direction, that must be provided by the tension in the string. Hence

T = \frac{m_{A}v^{2}}{L+\frac{d}{2}} \quad (9)

The speed of the block decreases with time due to friction, hence the speed of the block is maximum at the beginning of the motion, therfore the maximum tension is

T_{max} = \frac{m_{A}v_{0}^{2}}{L+\frac{d}{2}} \quad (10)

ii) The forces acting on the block are

a) Tension: Acting in the radially inwards direction, hence it is always perpendicular to the velocity of the block, therefore it does not change the speed of the block.

b) Friction: Acting tangentially, in the direction opposite to the velocity of the block at any given time, therefore it decreases the speed of the block.

The speed decreases linearly with time in the same manner as derived in part (C), using the expression for tension in part (D)(i) we can see that the tension in the string also decreases with time (in a quadratic manner to be specific).

8 0
3 years ago
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