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Pachacha [2.7K]

3 years ago

11

While loading, a truck moves 10 meters west from point A to point B in 5 seconds. Then it moves back to point A in 5 seconds. Wh

at are the average speed and average velocity of the truck?

1 answer:

Talja [164]3 years ago

5 0

Answer:

2m/s

Explanation:

distance(s)=10m

time taken(t)=5s

Now,

Speed(s)=s/t

=10m/5s

=2m/s [•.• 10/5=2]

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A football field is 100 yards long. If it takes a person 20 seconds to run its length, how fast were they running? Record your a

Black_prince [1.1K]

it is 5 yds/s to run the football firld

3 0

3 years ago

Physics question, please help me?

Damm [24]

As we know that block of steel is continue to be in moving state

so here the friction must be kinetic friction between two surface

so we know that formula of kinetic friction must be

$F_k = \mu_k mg$

now we have

$\mu_k = 0.57$

$mg = 25 N$

now from the above equation we have

$F_k = 0.57\times 25$

$F_k = 14.25 N$

So here we need atleast 14.25 N force to continue sliding the box now as per given options all forces which are less than 14.25 N is not correct

Hence correct answer must be 18.0 N

<u>D) 18.0 N</u>

7 0

4 years ago

A helicopter (m = 3250 kg) is cruising at a speed of 56.9 m/s atan altitude of 185 m. What is the total mechanical energy of the

taurus [48]

Answer:

The mechanical energy of the helicopter is $1.12\times 10^7\ J$ .

Explanation:

It is given that,

Mass of the helicopter, m = 3250 kg

Speed of the helicopter, v = 56.9 m/s

Position of the helicopter, h = 185 m

The energy possessed by an object due to its motion is called its kinetic energy. It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 3250\times (56.9)^2$

$E=5.26\times 10^6\ J$

The energy possessed by an object due to its position is called its potential energy. It is given by :

$E=mgh$

$E=3250\times 9.8\times 185$

$E=5.89\times 10^6\ J$

The sum of kinetic and potential energy is called mechanical energy of the system. It is given by :

$M=5.26\times 10^6+5.89\times 10^6$

$M=11.15\times 10^6\ J$

or

$M=1.12\times 10^7\ J$

So, the mechanical energy of the helicopter is $1.12\times 10^7\ J$ . Hence, this is the required solution.

6 0

4 years ago

Whats the answer to this question show in the picture 2 questions

Alexeev081 [22]

Answer: i think it would be watts

8 0

3 years ago

A projectile is fired from a height of 80 M above sea level, horizontally with a speed of 360 M / S, calculate: The time it take

Maslowich

Answer:

(a) The projectile takes approximately 4.420 seconds to reach the water, (b) The horizontal scope of the projectile is 1591.2 meters, (c) The remaining height to descend after 2 seconds of being launched is 63.624 meters.

Explanation:

The projectile experiments a parabolic motion, where horizontal speed remains constant and accelerates vertically due to the gravity effect. Let consider that drag can be neglected, so that kinematic equation are described below:

$x = x_{o}+v_{o,x} \cdot t$

$y = y_{o} + v_{o,y}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$x_{o}$ , $y_{o}$ - Initial horizontal and vertical position of the projectile, measured in meters.

$v_{o,x}$ , $v_{o,y}$ - Initial horizontal and vertical speed of the projectile, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in meters per square second.

$x$ , $y$ - Current horizontal and vertical position of the projectile, measured in meters.

Given that $x_{o} = 0\,m$ , $y_{o} = 80\,m$ , $v_{o,x} = 360\,\frac{m}{s}$ , $v_{o,y} = 0\,\frac{m}{s}$ and $g = -9.807\,\frac{m}{s^{2}}$ , the kinematic equations are, respectively:

$x = 360\cdot t$

$y = 80-4.094\cdot t^{2}$

(a) If $y = 0\,m$ , the time taken for the projectile to reach the water is:

$80 - 4.094\cdot t^{2} = 0$

$t = \sqrt{\frac{80}{4.094} }\,s$

$t \approx 4.420\,s$

The projectile takes approximately 4.420 seconds to reach the water.

(b) The horizontal scope is the horizontal distance done by the projectile before reaching the water. If $t \approx 4.420\,s$ , the horizontal scope of the projectile is:

$x = 360\cdot (4.420)$

$x = 1591.2\,m$

The horizontal scope of the projectile is 1591.2 meters.

(c) If $t = 2\,s$ , the height that remains to descend is:

$y = 80-4.094\cdot (2)^{2}$

$y = 63.624\,m$

The remaining height to descend after 2 seconds of being launched is 63.624 meters.

6 0

3 years ago