1answer.
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
Tatiana [17]
3 years ago
11

A speedboat increases its speed from 14.5 m/s to 29.3 m/s in a distance of 172 m.

Physics
1 answer:
dalvyx [7]3 years ago
6 0

Answer:

a. Acceleration, a = 1.88 m/s²

b. Time, t = 7.87 seconds.

Explanation:

Given the following data;

Initial velocity, U = 14.5m/s

Final velocity, V = 29.3m/s

Distance, S = 172m

a. To find the acceleration of the speedboat;

We would use the third equation of motion;

V² = U² + 2aS

Substituting into the formula

29.3² = 14.5² + 2a*172

858.49 = 210.25 + 344a

344a = 858.49 - 210.25

344a = 648.24

a = 648.24/344

Acceleration, a = 1.88 m/s²

b. To find the time;

We would use the first equation of motion;

V = U + at

29.3 = 14.5 + 1.88t

1.88t = 29.3 - 14.5

1.88t = 14.8

Time, t = 14.8/1.88

Time, t = 7.87 seconds.

You might be interested in
A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.
iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

F = \frac{kq_1q_2}{d^2}

Here

k = Coulomb's Constant

q_{1,2} = Charge of each object

d = Distance

Our values are given as,

q_1 = 1 \mu C

q_2 = 6 \mu C

d = 1 m

k =  9*10^9 Nm^2/C^2

a) The electric force on charge q_2 is

F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}

F_{12} = 54 mN

Force is positive i.e. repulsive

b) As the force exerted on q_2 will be equal to that act on q_1,

F_{21} = F_{12}

F_{21} = 54 mN

Force is positive i.e. repulsive

c) If q_2 = -6 \mu C, a negative sign will be introduced into the expression above i.e.

F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}

F_{12} = F_{21} = -54 mN

Force is negative i.e. attractive

6 0
3 years ago
Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

v_f = 1.034m/s

8 0
3 years ago
Four resistors , R1=12ohms, R2= 15 ohms , R3= 18 ohms and R4=10 ohms are connected to 6 v battery in series what is the total re
morpeh [17]

Answer:

55 ohms

Explanation:

if it was series circuit, then you just need to add all the resistances

6 0
3 years ago
What is the freezing point of an aqueous solution that boils at 101 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your an
astra-53 [7]

Answer:

-3.63 degree Celsius

Explanation:

We are given that

Boiling point of solution=T_b=101^{\circ} C

Boiling point water=100 degree Celsius

K_f=1.86K/m

K_b=0.512 K/m

\Delta T_b=T-T_0

Where T=Boiling point of solution

T_0=Boiling point of pure solvent

\Delta T_b=101-100=1^{\circ}C

\Delta T_b=k_bm

Using the formula

1=0.512\times m

Molality,m=\frac{1}{0.512} m

\Delta T_f=k_fm

Using the formula

\Delta T_f=\frac{1}{0.512}\times 1.86

\Delta T_f=3.63 C

We know that

\Delta T_f=T_0-T_1

Where T_0 =Freezing point of solvent

T_1= Freezing point of solution

Using the formula

3.63=0-T_1

Freezing point of water=0 degree Celsius

T_1=0-3.63=-3.63 C

Hence, the freezing point of solution=-3.63 degree Celsius

7 0
3 years ago
Ediment deposited where a river flows into an ocean or lake
Triss [41]
The answer is Delta.
3 0
3 years ago
Other questions:
  • How is a wavelength measured in a compressional wave?
    13·1 answer
  • An automobile traveling along a straight road
    6·1 answer
  • ANSWER ASAPPPPPPPP BE 100% CORRECT
    11·1 answer
  • What is the exact time of a full earth rotation?
    5·2 answers
  • A car of mass 1270 kg makes a 15.0 m radius turn at 11.8 m/s on flat ground. How much centripetal force acts on it?
    13·1 answer
  • Where on a roller coaster are there equal amounts of potential energy and kinetic energy?
    13·1 answer
  • If car is going at a speed of 80 km/h how far will it travel in two hours
    5·2 answers
  • The figure(Figure 1) shows the angular-velocity-versus-time graph for a particle moving in a circle, starting from θ0=0 rad at t
    12·1 answer
  • Can you explain why number 3 is correct?
    10·1 answer
  • Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!