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inessss [21]
3 years ago
7

It takes a bus driver 30 minutes to pick up students from four stops. The last stop is at the corner of Green Street and Route 7

0. The most direct route to school is north on Route 70. The speed limit on Route 70 is 35 kph. The bus driver wants to calculate the average velocity from the last stop to school.
Can this be done from the information given?


No, the distance from the last stop to the school and the time it takes to travel that distance are required.


No, the distance from the first stop to the school and the average speed are required.


Yes, if the average speed is 35 kph, and it takes 30 minutes, the average velocity is 17.5 kph.


Yes, if the speed limit is 35 kph, and the elapsed time is 30 minutes, the average velocity is 17.5 kph north
Physics
1 answer:
deff fn [24]3 years ago
7 0

Answer:

No, the distance from the last stop to the school and the time it takes to travel that distance are required.

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An object is moving east, and its velocity changes from 65 m/s to 25 m/s in 10 seconds Which describes the acceleration?
Rasek [7]

Answer:

4 m/s in negative acceleration

Explanation:

Acceleration = V- U/t

Where V is the final velocity

U is the initial velocity and t is the time given.

U = 65 m/s

V= 25 m/s

T= 10 seconds

Acceleration= (25m/s - 65m/s)÷10secs

= - 40/10

= -4m/s^2

Hence, it has a negative acceleration.

8 0
3 years ago
5. Find the mass of a car that is traveling at a velocity of 35 m/s West.
sattari [20]

Answer:

m = 9795.9 kg

Explanation:

v = 35 m/s

KE = 6,000,000 J

Plug those values into the following equation:

KE = \frac{1}{2} mv^{2}

6,000,000 J = (1/2)(35^2)m

---> m = 9795.9 kg

3 0
3 years ago
Waves can transfer energy through
lina2011 [118]

Answer:

electromagnetic waves

Explanation:

"wave" is a common term for a number different ways in which energy is transferred

3 0
3 years ago
The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about
nadya68 [22]

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s

a) The vertical speed when the player leaves the ground is 4.45 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s

Time taken to reach the maximum height is 0.45 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

6 0
3 years ago
A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force of friction on the cart
Ber [7]
Newton's second law states that the resultant of the forces applied to an object is equal to the product between the object's mass and its acceleration:
\sum F = ma
where in our problem, m is the mass the (child+cart) and a is the acceleration of the system.

We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force F_f acting in the opposite direction. So Newton's second law can be rewritten as
F-F_a = ma
or
F=ma + F_f

since the frictional force is 15 N and we want to achieve an acceleration of a=1.50 m/s^2, we can substitute these values to find what is the force the man needs:
F=(30 kg)(1.5 m/s^2)+15 N=60 N
8 0
3 years ago
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