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dimulka [17.4K]
2 years ago
11

Hola tengo un taller de física y nose como resolver este pregunta

Physics
1 answer:
Blizzard [7]2 years ago
8 0

a) 0.26 h

b) 71.4 km

Explanation:

a)

In order to solve the problem, we have to know what is the final velocity of the car.

Here, we assume that the final velocity reached by the car is

v=300 km/h

Therefore, we can find the time taken by the car to reach this velocity by using the suvat equation:

v=u+at

where:

u = 250 km/h is the initial velocity

a=190 km/h^2 is the acceleration of the car

v = 300 km/h is the final velocity

t is the time

Solving for t, we find:

t=\frac{v-u}{a}=\frac{300-250}{190}=0.26 h

b)

In order to find the distance covered by the car, we can use the following suvat equation:

s=ut+\frac{1}{2}at^2

where:

s is the distance covered

u is the initial velocity

a is the acceleration

t is the time

For the car in this problem, we have:

u = 250 km/h

t = 0.26 h (calculated in part a)

a=190 km/h^2

Therefore, the distance covered is

s=(250)(0.26)+\frac{1}{2}(190)(0.26)^2=71.4 km

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vazorg [7]

Answer:

10m/s^2

Explanation:

Given data

velocity= 40m/s

time= 4 seconds

Acceleration a =????

We know that

a= velocity/time

substitute

a=40/4

a= 10m/s^2

Hence the acceleration will be 10m/s^2

3 0
2 years ago
If the lily pads are spaced 2.4 m apart and Ferdinand jumps with a speed of 5 m/s taking 0.6 s to go from lily pad to lily pad,
Zina [86]

Answer:

36.87^{\circ}

Explanation:

v = Velocity of Ferdinand = 5 m/s

\theta = Angle of jump

T = Time taken = 0.6 s

R = Distance between lily pads = 2.4 m

Horizontal range is given by

R=v_xT\\\Rightarrow R=vcos\theta T\\\Rightarrow cos\theta=\dfrac{R}{vT}\\\Rightarrow \theta=cos^{-1}\dfrac{R}{vT}\\\Rightarrow \theta=cos^{-1}\dfrac{2.4}{5\times 0.6}\\\Rightarrow \theta=36.87^{\circ}

The angle at which Ferdinand make each of his jumps is 36.87^{\circ}

8 0
3 years ago
What is the mass of a truck in grams of it produces a force of 1500N while accelerating at a rate of 6 m/s²?​
aleksley [76]

Answer:

250,000

Explanation:

<h2> </h2>

<h2>formula = ( F=ma </h2>

  • F=1500N
  • a=6m/s^2
  • F= ma
  • m=?
  • 1500/6 = m
  • m=250 kg
  • 1kg =1000gm so 250kg =250,000gm
  • m =250×10^3 gm
5 0
3 years ago
The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the st
Aleonysh [2.5K]

Answer:

a = 5.83 \times 10^{-4} m/s^2

Explanation:

Since the system is in international space station

so here we can say that net force on the system is zero here

so Force by the astronaut on the space station = Force due to space station on boy

so here we know that

mass of boy = 70 kg

acceleration of boy = 1.50 m/s^2

now we know that

F = ma

F = 70(1.50) = 105 N

now for the space station will be same as above force

F = ma

105 = 1.8 \times 10^5 (a)

a = \frac{105}{1.8 \times 10^5}

a = 5.83 \times 10^{-4} m/s^2

3 0
3 years ago
If there are 50 people per square kilometer in a city, and the area of the city is 1.5 × 10 square kilometers. What is the total
qaws [65]

Answer:

750 people

Explanation:

From the question,

Number of people in the city = population density×Area of the city

N = D×A.......................... Equagtion 1

Where N = Number of people in the city, D = population density, A = Area of the city.

Given: D = 50 people per square kilometer, A = 1.5×10 square kilometer.

Substitute into equation 1

N = 50(1.5×10)

N = 750 people.

Hence the total number of people in the city is 750 people.

6 0
3 years ago
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