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10_Phys U1 Summative (45MC)
By Lester Orellana
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Last updated about 6 years ago
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Acceleration is ...
the rate of change of velocity
the change in time over velocity
the change in speed over time
the rate of change of speed
The formula for average speed is
total acceleration divided by time
total velocity divided by time
total displacement divided by time
total distance divided by time
In velocity time graphs, the area under the line represents ...
acceleration
change in displacement
change in speed
change in acceleration
In displacement-time graphs, the gradient of the line represents...
change in velocity
change in acceleration
speed
velocity
acceleration
The magnitude acceleration due to Earth's gravity is
9.81 m/s^2
-9.81 ms^2
-9.81 m/s^2
9.81 ms^2
The 1st Newton's Law states that every object remains resting or moving with constant ___________ until external __________ is applied
velocity, mass
speed, mass
speed, force
velocity, force
The main reason why we don't have acceleration according to one of the Newton's Law is...
The vector addition of all the forces related to object is zero
The vector addition of the forces on the vertical is zero
The vector addition of the forces on the horizontal is zero
The terminal velocity is reached when ...
there is air resistance and magnitude of the mass equals the weight.
there is air resistance and magnitude of the force equals the weight.
there is air resistance and magnitude of the friction equals the weight.
Is the measurement of the resistance of an object to change its state of motion. (start moving or stopping)
Inertia
Friction in general
Static Friciton
Dynamic Friction
When an unbalanced force moves the object with acceleration is part of ...
2nd Newton's Law
1st Newton's Law
3rd Newton's Law
Is the force that is opposed to the motion or to start moving.
Static Friction
Friction
Dynamic Friction
Normal Force
Is the force that is perpendicular to the surface and is related to friction.
Static Friction
Friction
Normal Force
Dynamic Friction
Choose the only true statement.
Stopping distance is the addition of the reaction distance and braking distance.
Stopping distance is the addition of all the braking distance.
Reaction distance is the addition of the stopping distance and braking distance.
Braking distance is the addition of the reaction distance and stopping distance.
The formula for Earth's gravitational force is ...
W = mg
F = mg
W = ma
F = ma
Mass and weight ....
Are the same because both are measured in Newtons.
Are the same because both are measured in kilograms.
are not the same because mass is always constant.
are not the same because weight is always constant.
Third Newton's Law says that for every actions there is a reaction in ...
same direction but opposite in magnitude.
same magnitude but opposite in direction.
A vector is a physical quanity with
direction
mass
unit
The letter "u" represents
initial velocity or initial speed
initial speed
initial velocity
The unit for gravity is
ms
-2
m/s
-2
ms
2
In a projectile motion, the horizontal velocity is _______________.
changing due to gravity 9.81
constant because there is no air resistance affecting the projectile.
constant because mass is constant.
changing due to acceleration of 9.81
If velocity 28.28 m/s, find the horizontal velocity
= 28.28 x cos25.6ยบ = 25.51 m/s
= 28.28 x sin25.6ยบ = 12.22 m/s
= 28.28 x cos25.6ยบ = 12.22 m/s
= 28.28 x sin25.6ยบ = 25.51 m/s
If velocity 28.28 m/s, to find the highest altitude ...
consider v = 28.28 x sin25.6 in vertical "suvat"
consider v = zero in horizontal "suvat"
consider v = 28.28 x sin25.6 in horizontal "suvat"
consider v = zero in vertical "suvat"
If velocity 28.28 m/s, to find the range we should consider ...
The velocity is the cos of 25.6ยบ
The velocity is the sin of 25.6ยบ
the time is exactly what takes to reach the highest altitude.
the time is double of what takes to reach the highest altitude.
The final velocity reaching the starting point is
20 m/s
20 x cos90ยบ
-20 m/s
-20 x cos90ยบ
The acceleration in this motion is
9.81 m/s^2
-9.81 m/s^-2
-9.81 m/s^2
9.81 m/s^-2
A car is traveling at +20m/s for 30 seconds, then returns decreasing the velocity until it stops in 20 seconds.
Find the total displacement.
+400 m
+800 m
+600 m
+1,000 m
This displacement-time graph represents...
negative acceleration when velocity starts in the negative region.
positive acceleration when velocity starts in the negative region.
negative acceleration when velocity starts in the positive region.
positive acceleration when velocity starts in the positive region.
This displacement time graph represents
positve acceleration and positive constant velocity
no acceleration and negative constant velocity
no acceleration and positive constant velocity
positve acceleration and negative constant velocity
This velocity time graph represents
zero acceleration
negative acceleration
positive acceleration
Choose the only true statement
This is velocity-time graph when acceleration is negative
This is displacement time graph when acceleration is negative and velocity is constant.
This is displacement time graph when acceleration is positive and velocity is constant.
This is velocity-time graph when acceleration is positive
This is the area when the object is moving with acceleration zero.
C
B
A
This is the area when use ยต
d
C
A
B
What is the velocity when t=10s
25 m/s
50 m/s
75 m/s
100 m/s
What is the net force if the weight is 10,000 N? (g = 10)
30,000 N
300 N
3,000 N
If the net force is -400 N in this problem, How much is the applied force?
-1000 N
900 N
1,000 N
-900 N
How much is the weight of the ball? (g=10)
300
10
30
100
How much is the weight of the crate? (g=10)
60
5
500
50
How much is the mass of the crate? (g=10)
5 kg
50 kg
15 kg
10.5 kg
How much is the maximum static friction of the crate? (g=10)
30 N
26.4 N
22 N
50.44 N
2.2 N
What is the force of impact of the ball on the crate? (g=10)
22 N
13 N
4.4 N
30 N
Is it going to be moved after the impact of the ball on the crate? (g=10)
No, because impact force is smaller than maximum coefficient static friction.
No, because impact force is smaller than maximum static friction.
Yes, because impact force is greater than maximum coefficient static friction.
Yes, because impact force is greater than maximum static friction.
If g=10. How much is the normal force?
-10,000 N
-1,000 N
+1,000 N
+10,000 N
If g=10. How much is the friction force is net force is 3,000 N?
3,000 N
30,000 N
-30,000 N
-3,000 N
If g=10. How much is the net force is frictional forces are 4,000 N?
+14,000 N
-6,000 N
-14,000 N
+6,000 N
If g=10. How much is the mass if frictional forces are -2,000 N?
8 kg
12 kg
18 kg
2 kg