answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ivan
2 years ago
8

A car with an initial velocity of 16.0 meters per second east slows uniformly to 6.0 meters per second east in 4.0 seconds. What

is the acceleration of the car during this 4.0-second interval?
Physics
1 answer:
pychu [463]2 years ago
5 0
(6-16)/4.0=-2.5 m/s²
Acceleration of the car is -2.5 m/s²
You might be interested in
A raft is made of a plastic block with a density of 650 kg/m 3 , and its dimensions are 2.00 m à 3.00 m à 5.00 m. 1. what is the
cupoosta [38]
1) The volume of the raft is the product between the lenghts of its three dimensions:
V = (2.00 m)(3.00m)(5.00m)=30 m^3

2) The mass of the raft is the product between its density, d, and its volume, V:
m=dV=(650 kg/m^3)(30 m^3)=19500 kg

3) The weight of the raft is the product between its mass m and the gravitational acceleration, g=9.81 m/s^2:
W=mg=(19500 kg)(9.81 m/s^2)=1.91 \cdot 10^5 N

4) The apparent weight is equal to the difference between the weight of the raft and the buoyancy (the weight of the displaced fluid):
W_a = W- \rho_W V_{disp} g
where \rho _W = 1000 kg/m^3 is the water density and V_{disp} is the volume of displaced fluid.
The density of the raft (650 kg/m^3) is smaller than the water density (1000 kg/m^3), this means that initially the buoyancy (which has upward direction) is larger than the weight (downward direction) and so the raft is pushed upward, until it reaches a condition of equilibrium and it floats. At equilibrium, the weight and the buoyancy are equal and opposite in sign:
W=B=\rho _W V_{disp} g
and therefore, the apparent weight will be zero:
W_a = W-B=W-W=0

5) The buoyant force B is the weight of the displaced fluid, as said in step 4):
B=\rho_W V_{disp} g
When the raft is completely immersed in the water, the volume of fluid displaced V_{disp} is equal to the volume of the raft, V_{disp}=V. Therefore the buoyancy in this situation is
B= \rho_W V g = (1000 kg/m^3)(30 m^3)(9.81 m/s^2)=2.94 \cdot 10^5 N
However, as we said in point 4), the raft is pushed upward until it reaches equilibrium and it floats. At equilibrium, the buoyancy will be equal to the weight of the raft (because the raft is in equilibrium), so:
B=W=1.91 \cdot 10^5 N

6) At equilibrium, the mass of the displaced water is equal to the mass of the object. In fact, at equilibrium we have W=B, and this can be rewritten as
mg = m_{disp} g
where m_{disp}= \rho_W V_{disp} is the mass of the displaced water. From the previous equation, we obtain that m_{disp}=m=19500 kg.

7) Since we know that the mass of displaced water is equal to the mass of the raft, using the relationship m=dV we can rewrite m=m_{disp} as:
d V =d_W V_{disp}
and so
V_{disp}= \frac{d V}{d_W}= \frac{(650 kg/m^3)(30m^3)}{1000kg/m^3}= 19.5 m^3

8) The volume of water displaced is (point 7) 19.5 m^3. This volume is now "filled" with part of the volume of the raft, therefore 19.5 m^3 is also the volume of the raft below the water level. We can calculate the fraction of raft's volume below water level, with respect to the total volume of the raft, 30 m^3:
\frac{19.5 m^3}{30 m^3}\cdot 100= 65 \%
Viceversa, the volume of raft above the water level is 30 m^3-19.5 m^3 = 10.5 m^3. Therefore, the fraction of volume of the raft above water level is
\frac{10.5 m^3}{30 m^3}\cdot 100 = 35 \%

9) Let's repeat steps 5-8 replacing \rho _W, the water density, with \rho_E=806 kg/m^3, the ethanol density.

9-5) The buoyant force is given by:
B=\rho _E V_{disp} g = (806 kg/m^3)(30 m^3)(9.81 m/s^2)=2.37 \cdot 10^5 N
when the raft is completely submerged. Then it goes upward until it reaches equilibrium and it floats: in this condition, B=W, so the buoyancy is equal to the weight of the raft.

9-6) Similarly as in point 6), the mass of the displaced ethanol is equal to the mass of the raft: 
m_E = m = 19500 kg

9-7) Using the relationship d= \frac{m}{V}, we can find the volume of displaced ethanol:
V_E =  \frac{m}{d_E} = \frac{19500 kg}{806 kg/m^3}=24.2 m^3

9-8) The volume of raft below the ethanol level is equal to the volume of ethanol displaced: 24.2 m^3. Therefore, the fraction of raft's volume below the ethanol level is 
\frac{24.2 m^3}{30 m^3}\cdot 100 = 81 \%
Consequently, the raft's volume above the ethanol level is 
30 m^3 - 24.2 m^3 = 5.8 m^3
and the fraction of volume above the ethanol level is
\frac{5.8 m^3}{30 m^3}\cdot 100 = 19 \%
8 0
2 years ago
. A horizontal steel spring has a spring constant of 40.0 N/m. What force must be applied to the spring in order to compress it
ANEK [815]

Ans    4 more to be exact

Explanation:

7 0
1 year ago
dopasuj wartości pracy z ramki do przedstawionych sytuacji ,a nastepnie wyraż te prace w dżulach uwaga jedna wartośc pracy nie b
Eduardwww [97]

Answer:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Explanation:

Całą część pytania można znaleźć na poniższym schemacie.

Z diagramu załączonego poniżej; mamy

A - samolot lotniczy

B - dźwig

C - podnoszenie ciężarów

Wszyscy to wiemy ;

1kJ = 1000 J

1MJ = 1000000 J

Mamy cztery opcje; i.e 200 MJ, 800 kJ, 1.6 kJ  and 250 mJ

Z czterech opcji można wykluczyć 250 mJ, ponieważ jest to 0,25 J, co przedstawia bardzo niską energię w porównaniu z trzema warunkami pokazanymi na schemacie.

Więc:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Największą pracę wykona samolot. Jest tak, ponieważ ma bardzo dużą masę i bardzo dużą prędkość. W związku z tym istnieje potrzeba wytworzenia ogromnej ilości ciepła i energii.

Z drugiej strony żuraw może podnieść ładunek o wiele większy i przewyższa ciężar ciężaru, więc praca wykonywana przez dźwig musi być zdecydowanie większa niż praca ciężarka.

3 0
2 years ago
At the instant a ball is thrown horizontally with a large force, an identical ball is dropped from the same height. which ball h
sergejj [24]
The one that was dropped
8 0
2 years ago
Read 2 more answers
How much force is required to pull a spring 3.0 cm from
avanturin [10]

Answer:

I know that T= kx where T is the tension which equaka the force og gravity = mg = 1.37 * 10 = 13.7 x is the elongation of the spring so the length after dangling the object minus the original length.

I hope it helps

plz let me know if it is wrong or right.

4 0
2 years ago
Other questions:
  • A 0.0010-kg pellet is fired at a speed of 50.0m/s at a motionless 0.35-kg piece of balsa wood. When the 
    6·2 answers
  • While dragging a crate a workman exerts a force of 628 N. Later, the mass of the crate is increased by a factor of 3.8. If the w
    10·2 answers
  • Avi, a gymnast, weighs 40 kg. She is jumping on a trampoline that has a spring constant value of 176,400 . If she compresses the
    12·2 answers
  • Mt. Asama, Japan, is an active volcano complex. In 2009, an eruption threw solid volcanic rocks that landed far from the crater.
    10·1 answer
  • A mirror forms an erect image 40cm from the object and one third its height where must the mirror be situated ​
    12·1 answer
  • To practice Problem-Solving Strategy 25.1 Power and Energy in Circuits. A device for heating a cup of water in a car connects to
    5·1 answer
  • A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i
    10·1 answer
  • Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x =
    11·1 answer
  • PLEASE HELPPP 100 POINTS HURRY !!!!Which diagram best illustrates the magnetic field of a bar magnet? A bar magnet with a north
    13·2 answers
  • Which option is part of designing a set of experimental procedures?
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!