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Usimov [2.4K]
3 years ago
9

A school bus takes 0.53 hours to reach the school from your house. If the average speed of the bus is 19km/h, what is the displa

cement of the bus during the trip?
Physics
2 answers:
Artist 52 [7]3 years ago
5 0
D = v*t= 19*0,53 = 10.07 km
Harlamova29_29 [7]3 years ago
4 0

Answer:

The displacement of the bus during the trip is 10.07 km.

Explanation:

Given that,

Time = 0.53 h

Average speed of bus = 19 km/h

We need to calculate the displacement of the bus

Using formula of average velocity

v_{av}=\dfrac{D}{t}

D = v_{avg}\times t

Where, D = displacement

t = time

Put the value into the formula

D=19\times0.53

D=10.07\ km

Hence, The displacement of the bus during the trip is 10.07 km.

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A train is traveling at a speed of 80\,\dfrac{\text{km}}{\text{h}}80 h km ​ 80, start fraction, start text, k, m, end text, divi
trasher [3.6K]

Answer:

The distance the train travels before coming to a (complete) stop = 40/81 km which is approximately 493.83 meters

Explanation:

The initial speed of the train u = 80 km/h = 22 2/9 m/s = 22.\bar 2 m/s

The magnitude of the constant acceleration with which the train slows, a = 0.5 m/s²

Therefore, we have the following suitable kinematic equation of motion;

v² = u² - 2 × a × s

Where;

v = The final velocity = 0 (The train comes to a stop)

s = The distance the train travels before coming to a stop

Substituting the  values gives;

0² = 22.\bar 2² - 2 × 0.5 × s

2 × 0.5 × s = 22.\bar 2²

s = 22.\bar 2²/1 = 493 67/81 m = 40/81 km

The distance the train travels before coming to a (complete) stop = 40/81 km ≈ 493.83 m.

7 0
3 years ago
Read 2 more answers
En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a
Maru [420]

Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

Q=mc(T_2-T_1)

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

Q_1=-Q_2

for water we have to

c = 4.18J / g ° C

replacing we have

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

Q=mc(T_2-T_1)

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

Q_1=-Q_2

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

7 0
3 years ago
At what temperature do the fahrenheit and celsius scales give the same reading?
anygoal [31]
At -40.

-40 gives the same reading for Fahrenheit and Celsius scale.
7 0
3 years ago
Can someone explain to how to calculate this
Karo-lina-s [1.5K]

answer

option d is the correct answer

explanation

as we know frequency is equal to 1 /t

f= 457 Hz

t=1

SO, 1/457

=0.0022sev

3 0
3 years ago
A spring is hung from the ceiling. A 0.442-kg block is then attached to the free end of the spring. When released from rest, the
siniylev [52]

Answer:

K=58.8N/m

Explanation:

From the question we are told that:

Mass M=0.442

Drop distance d=0.150

Generally the equation for Spring Constant is mathematically given by

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 K=\frac{2*0.442*9.8}{1.150}

 K=58.8N/m

6 0
3 years ago
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