All categories

3 years ago

If a bus travels 120 miles and travels at 35 miles per hour, how long will the trip take?

Physics

1 answer:

Artyom0805 [142]3 years ago

7 0

Speed = Distance/Time

we have speed and distance from the question;

Now,

35 = 120/t

35/120 = 1/t

120/35 = t

3.42 hours = t

You might be interested in

A test car starts from rest on a horizontal circular track of 115-m radius and increases its speed at a uniform rate to reach 90

Wewaii [24]

Answer:

a= 3.49 m/s^2

Explanation:

magnitude of total acceleration = sqrt{radial acceleration^2+tangential acceleration^2}.

we know that tangential acceleration a_t= change in velocity /time taken

now 90 km/h = 25 m/s

a_t = 25/17 = 1.47 m/s^2.

radial acceleration a_r = v^2/r

v= a_t×t = 1.47×13 = 19.11 m/s

a_r = 19.11^2/115= 3.175

now,

$a= \sqrt{a_t^2+a_r^2}$

$a= \sqrt{1.47^2+3.175^2}$

a= 3.49 m/s^2

3 0

3 years ago

What is the total current flowing through a 45 ohm resistor (1.4 amps) and a 20 ohm resistor (1.8 amps)?

BigorU [14]

Send me the circuit diagram

7 0

3 years ago

The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car's moti

Ksenya-84 [330]

Answer: The answer is B

Explanation: It is staying in a steady speed position

5 0

2 years ago

Read 2 more answers

What is the molar mass of an ideal gas if a 0.800 g sample of this gas occupies a volume of 200. mL at 50.0 oC and 720. mm Hg?

Paladinen [302]

PV=nRT
(720/760)(0.200)=(0.800/x)(0.08206)(323.15)
(0.1894736842)=(0.800/x)(0.08206)(323.15)
.0071451809=(0.800/x)
x=MM=111.9635758 g/mol

7 0

3 years ago

Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou

kirza4 [7]

Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

i = current flow = 3 A
t = time interval for which the current flow = $4\ h = 4\times 3600\ s = 14400\ s$
V = terminal voltage of the battery
R = rate of energy = 9 cents/kWh

<u>Assume:</u>

Q = charge transported as a result of charging
E = energy expended
C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

$\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\$

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

$E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\$

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

$\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents$

Hence, 0.108V cents is the charging cost of the battery.

4 0

3 years ago