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3 years ago

A train runs from New Delhi to Hyderabad it covers first of 420 km in 7 hours and the next distance of 360 km in 6 hours

Physics

2 answers:

Mkey [24]3 years ago

5 0

Explanation:

the answer is 16.7 miles or 60kmph

maks197457 [2]3 years ago

5 0

Answer:

II. Gopi takes part in a car race. He drives a distance of 70 km each in the first, second and third hours.

Which of the following statements is true.

a) I, is an example uniform motion and II is an example of non-uniform motion.

b) I is an example of nonuniform motion and II is an example of uniform motion.

c) I and II are examples of uniform motion.

d) I and II are examples of nonuniform motion.

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A 75.0-kg man steps off a platform 3.10 m above the ground. he keeps his legs straight as he falls, but his knees begin to bend

sasho [114]

Refer to the diagram shown below.

u = 0, the initial vertical velocity
Assume g = 9.8 m/s² and ignore air resistance.

At the first stage of landing on the ground, the distance traveled is
h = 3.1 - 0.6 = 2.5 m.
If v = the vertical velocity at this stage, then
v² = u² + 2gh
v² = 2*(9.8 m/s²)*(2.5 m) = 49 (m/s)²
v = 7 m/s

At the second stage of landing on the ground, let a = the acceleration (actually deceleration) that his body provides to come to rest.
The distance traveled is 0.6 m.
Therefore
0 = (7 m/s)² + 2(a m/s²)*(0.6 m)
a = - 49/1.2 = - 40.833 m/s²

Answers:
(a) The velocity when the man first touches the ground is 7.0 m/s.
(b) The acceleration is -40.83 m/s² (deceleration of 40.83 m/s²) to come to rest within 0.6 m.

8 0

3 years ago

The acceleration of a particle traveling along a straight line isa=14s1/2m/s2, wheresis in meters. Ifv= 0,s= 1 m whent= 0, deter

Yuliya22 [10]

Answer:

0.78m/s

Explanation:

We are given that

Acceleration= $a=\frac{1}{4}s^{\frac{1}{2}}m/s^2$

v=0, s=1 when t=0

We have to find the particle's velocity at s=2m

We know that

$a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=\frac{dv}{ds}v$

$vdv=ads$

$\int_{0}^{v} vdv=\int_{1}^{s}0.25s^{\frac{1}{2}}ds$

$\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2})^{s}_{1}$

By using formula: $\int x^ndx=\frac{x^{n+1}}{n+1}+C$

$\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2}}-1$

Substitute s=2

$\frac{v^2}{2}=\frac{0.50}{3}((2^{1.5})-1)$

$\frac{v^2}{2}=\frac{0.50}{3}\times 1.83$

$v^2=2\times 0.305=0.61$

$v=\sqrt{0.61}=0.78m/s$

Hence, the velocity of particle at s=2m=0.78m/s

3 0

3 years ago

Physics help please..............

adoni [48]

Answer:

3.68 m/s

Explanation:

Full answer in the attached picture

7 0

3 years ago

A bucket filled with water has a mass of 70kg and is hanging from a rope that is wound around a 0.054 m radius stationary cylind

guajiro [1.7K]

Answer:

T = 37.08 [N*m]

Explanation:

We must remember that torque is defined as the product of a force by a distance. This distance is measured from the point of application of force to the center of rotation of the rotating body.

The force is equal to the product of mass by gravitational acceleration.

$F=m*g\\F=70*9.81\\F=686.7[N]$

Now the torque can be calculated:

$T=F*r\\T=686.7*0.054\\T=37.08[N*m]$

3 0

2 years ago

A jet liner must reach a speed of 82 m/s for takeoff. If the

SIZIF [17.4K]

Answer:

The acceleration that the jet liner that must have is 2.241 meters per square second.

Explanation:

Let suppose that the jet liner accelerates uniformly. From statement we know the initial ( $v_{o}$ ) and final speeds ( $v_{f}$ ), measured in meters per second, of the aircraft and likewise the runway length ( $d$ ), measured in meters. The following kinematic equation is used to calculate the minimum acceleration needed ( $a$ ), measured in meters per square second:

$a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot d}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 82\,\frac{m}{s}$ and $d = 1500\,m$ , then the acceleration that the jet must have is:

$a = \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (1500\,m)}$

$a = 2.241\,\frac{m}{s^{2}}$

The acceleration that the jet liner that must have is 2.241 meters per square second.

3 0

2 years ago

A train runs from New Delhi to Hyderabad it covers first of 420 km in 7 hours and the next distance of 360 km in 6 hours​

A train runs from New Delhi to Hyderabad it covers first of 420 km in 7 hours and the next distance of 360 km in 6 hours