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SVEN [57.7K]
2 years ago
9

A car is traveling in a race.The car went from initial velocity of 35m/s to the final velocity of 65m/s in 5 seconds what was th

e acceleration
Physics
2 answers:
I am Lyosha [343]2 years ago
8 0
Acceleration is the change in velocity divided by time. The change in velocity is -30m/s and time is 5s. If you divide -30m/s by 5s, you get -6m/s<span>².</span>
OverLord2011 [107]2 years ago
6 0

Answer:

Acceleration, a=6\ m/s^2

Explanation:

Given that,

Initial speed of the car, u = 35 m/s

Final speed of the car, v = 65 m/s

Time taken, t = 5 s

We need to find the acceleration of the car. It is given by :

a=\dfrac{v-u}{t}

a=\dfrac{65-35}{5}

a=6\ m/s^2

So, the acceleration of the car is 6\ m/s^2. Hence, this is the required solution.

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What would the speed of each particle be if it had the same wavelength as a photon of yellow light (????=575.0 nm)? Proton (mass
PilotLPTM [1.2K]

Answer:

Proton: v=0.689 m/s

Neutron: v=0.688 m/s

Electron: v=1265.078 m/s

Alpha particle: v=0.173 m/s

Explanation:

De Broglie equation allows you to calculate the “wavelength” of an electron or any other particle or object of mass m that moves with velocity v:

λ=\frac{h}{mv}

h is the Planck constant: 6.626×10⁻³⁴\frac{kg.m^2}{s}

We know that the wavelength of the particle is 575 nm (575×10⁻⁹m), so we find the velocity v for each particle:

λ=\frac{h}{mv}

v=h÷(mλ)

<u>Proton:</u>

m=1.673×10⁻²⁴ g · \frac{1kg}{1000g}=1.673×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴\frac{kg.m^2}{s}÷(1.673×10⁻²⁷ kg×575×10⁻⁹m)

v=0.689 m/s

<u>Neutron:</u>

m=1.675×10⁻²⁴ g · \frac{1kg}{1000g}=1.675×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴\frac{kg.m^2}{s}÷(1.675×10⁻²⁷ kg×575×10⁻⁹m)

v=0.688 m/s

<u>Electron:</u>

m= 9.109×10⁻²⁸ g · \frac{1kg}{1000g}=9.109×10⁻³¹ kg

v=h÷(mλ)

v=6.626×10⁻³⁴\frac{kg.m^2}{s}÷(9.109×10⁻³¹ kg×575×10⁻⁹m)

v=1265.078 m/s

<u>Alpha particle:</u>

m=6.645×10⁻²⁴ g · \frac{1kg}{1000g}=6.645×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴\frac{kg.m^2}{s}÷(6.645×10⁻²⁷ kg×575×10⁻⁹m)

v=0.173 m/s

3 0
2 years ago
Read 2 more answers
Before leaving the house in the morning, you plop some stew in your slow cooker and turn it on Low. The slow cooker has a 160 Oh
guajiro [1.7K]

Answer:

Total charge flow through the cooker is 21600 C

Explanation:

As we know that the current flow through the cooker is given by Ohm's law

here it is given as

V = i R

i = \frac{V}{R}

i = \frac{120}{160}

i = \frac{3}{4} A

now the charge flow through it is given as

Q = i t

total time is t = 8 hours

Q = \frac{3}{4}(8 \times 60 \times 60)

Q = 21600 C

7 0
2 years ago
The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is c
puteri [66]

Answer:

64.59kpa

Explanation:

See attachment

6 0
2 years ago
While traveling from Boston to Hartford, Person A drives at a constant speed of 55 mph for the entire trip. Person B drives at 6
Ne4ueva [31]

Answer:

B will take 1.034 times the time of A from Boston to Hartford.

Explanation:

Let the distance from Boston to Hartford be S.

Person A drives at a constant speed of 55 mph for the entire trip,

Time taken by person A

             t_A=\frac{S}{55}

Person B drives at 65 mph for half the distance and then drives 45 mph for the second half of the distance.

Time taken by person B

            t_B=\frac{\frac{S}{2}}{65}+\frac{\frac{S}{2}}{45}=\frac{S}{130}+\frac{S}{90}=\frac{220S}{130\times 90}=\frac{11S}{585}

Ratio of time of arrival of B to A

                      \frac{t_B}{t_A}=\frac{\frac{11S}{585}}{\frac{S}{55}}=\frac{121}{117}=1.034

B will take 1.034 times the time of A from Boston to Hartford.

8 0
2 years ago
A negative charge of -0.550 μC exerts an upward 0.900-N force on an unknown charge that is located 0.300 m directly below the fi
allochka39001 [22]

Answer:

a.   q2 = 16.4μC, positive charge

b.   F = 0.900N

c.   downward

Explanation:

a. In order to calculate the charge of the unknown charge you use the following formula, for the electric force between two charges:

F_e=k\frac{q_1q_2}{r^2}            (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r: distance between the charges = 0.300m

q1: charge 1 = -0.550 μC = 0.550*10^-6C

q2: charge 2 = ?

Fe: electric force = 0.900N

The force exerted in the second charge points upward, then, the sign of the second charge is positive because this charge is getting closer to the first one.

You solve the equation (1) for the second charge ans replace the values of the other parameters:

q_2=\frac{r^2F_e}{kq_1}=\frac{(0.300m)^2(0.900N)}{(8.98*10^9Nm^2/C^2)(0.550*10^{-6}C)}\\\\q_2=1.64*10^{-5}C\\\\q_2=16.4*10^{-6}C=16.4*10\mu C

The values of the second charge is 1.64 μC

b. By the third Newton Law, you have that the force exerted in the second charge is equal to the force exerted by the first charge on the second one.

The force exerted on the first charge is 0.900N

c. The charges are attracting between them, then, the force exerted on the first charge is pointing downward.

3 0
2 years ago
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