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Norma-Jean [14]
3 years ago
10

How many seconds in a day? A hole day.

Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
4 0
1 year<span> consists of 365 days. 1 day has 24 hours, each hour has 60 minutes and each minute has 60 </span>seconds. <span>1 day = (24 hours/day) × (60 minutes/hour) × (60 seconds/minute) = 86400 seconds/day

Hope that helped :)</span>
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You are traveling on an airplane. The velocity of the plane with respect to the air is 110.0 m/s due east. The velocity of the a
Mice21 [21]
1. Vpa = 180m/s. @ 0 deg.
  Vag = 40m/s @ 120 deg,CCW.


<span> Vpg = Vpa + Vag,
 Vpg = (180 + 40cos120) + i40sin120,
  Vpg = 160 + i34.64,
 Vpg=sqrt((160)^2 + (34.64)^2)=163.7m/s.
</span>
<span>2. tanA = Y / X = 34.64 / 160 = 0.2165,
  A = 12.2 deg,CCW. = 12.2deg. North of East. </span>

3.  1 hr = 3600s. <span>d = Vt = 163.7m/s * 3600s = 589,320m.

hope this helps</span>
8 0
2 years ago
A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t
Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

                       v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time  

                       V = at + U

using equation  v - u = at to get line equation for the graph of the motion of the train on the incline plane

                       V_{x} = mt + V_{o}      where m is the slope

Comparing equation (1) and (2)

V = V_{x}

a = m    

U = V_{o}

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

          a = -  1.40 m/s²

Sunstituting a = -  1.40 m/s² and  u = 22 m/s

                        V_{x} = -1.40t + 22

                            V_{x} = -1.40(7.30) + 22

                             V_{x} = -10.22 + 22

                             V_{x} = 11. 78 m/s

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

           Area of triangle +  Area of rectangle

          [\frac{1}{2} * (22 - 11.78) * (7.30)]  + [(11.78 - 0) * (7.30)]

                           = 37.303 + 85.994

                           = 123. 297 m

                           ≈ 123. 30 m

                 

4 0
2 years ago
Read 2 more answers
Describe an experiment to show that pressure increases with the decrease in the area of surface
cluponka [151]
Answer:
press a baloon against one pin it bursts
but now arrange lot of pins parallel closely to each other if u press a baloon against them it does not burst hope this helps u
7 0
2 years ago
What are the set of equations called that can be used to quantify motion in the case of uniform acceleration?
KATRIN_1 [288]
The kinematic equations are used to <span>quantify motion in the case of uniform acceleration.

The other name is : SUVAT equations, where the letters signify: 
 displacement (s),
initial velocity (u),
 final velocity (v),
acceleration (a), and
 time (t).

There are three equations are attached in the picture: </span>

7 0
3 years ago
A very hard rubber ball (m = 0.5 kg) is falling vertically at 4 m/s just before it bounces on the floor. The ball rebounds back
saw5 [17]

Answer:

The force exerted by the floor is 80 N.

Explanation:

Given that,

Mass of ball = 0.5 kg

Velocity= 4 m/s

Time t = 0.05 s

When the ball rebounds then the kinetic energy is

K.E =\dfrac{1}{2}mv^2

Where, m = mass of ball

v = velocity of ball

Put the value into the formula

K.E=\dfrac{1}{2}\times0.5\times(4)^2

K.E = 4\ J

The average force exerted by the floor on the ball = change in kinetic energy over collision time

F = \dfrac{4}{0.05}

F=80\ N

Hence, The force exerted by the floor is 80 N.

4 0
2 years ago
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