Help me please i dont understand

Which type of physical activity is being performed in the picture?

mafiozo [28]

B strength training I think that’s the answer

3 0

3 years ago

An unmarked police car traveling a constant 95 km/h is passed by a speeder. Precisely 1.00 s after the speeder passes, the polic

Montano1993 [528]

Answer:

Speed of the speeder will be 28 m/sec

Explanation:

In first case police car is traveling with a speed of 90 km/hr

We can change 90 km/hr in m/sec

So $90km/hr=\frac{90\times 1000m}{3600sec}=25m/sec$

Car is traveling for 1 sec with a constant speed so distance traveled in 1 sec = 25×1 = 25 m

After that car is accelerating with $a=2m/sec^2$ for 7 sec

So distance traveled by car in these 7 sec

$S=ut+\frac{1}{2}at^2=25\times 7+\frac{1}{2}\times 2\times 7^2=224m$

So total distance traveled by police car = 224 m

This distance is also same for speeder

Now let speeder is moving with constant velocity v

so $224=\left ( 1+7 \right )v$

v = 28 m/sec

4 0

4 years ago

Read 2 more answers

How does gravitational pull affect planets with the same mass but different distance from the sun?

Yakvenalex [24]

Answer:

it just pulls them at the same time

Explanation:

5 0

4 years ago

A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$0=(64ft/s)^2-2(32ft/s^2)(y-960ft)$

$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

$v=-256ft/s$

5 0

3 years ago

Does the air exert a buoyant force on all objects in air or only on objects such as balloons that are very light for their size?

Citrus2011 [14]

Answer:

See explanation

Explanation:

Solution:-

Buoyancy is the force that causes objects to float. It is the force exerted on an object that is partly or wholly immersed in a fluid. Buoyancy is caused by the differences in pressure acting on opposite sides of an object immersed in a static fluid. It is also known as the buoyant force. Buoyancy is the phenomena due to Buoyant Force.

It is as an upward force exerted by a fluid that opposes the weight of an object immersed in a fluid. As we know, the pressure in a fluid column increases with depth. Thus, the pressure at the bottom of an object submerged in the fluid is greater than that at the top. The difference in this pressure results in a net upward force on the object which we define as buoyancy.

- The formula for buoyant force (Fb) is given:

Fb = ρ*g*V

- The force acts on all objects. However, it depends on the fluid density and amount of volume displaced.

- The Buoyant force exerted by air with density = 1.225 kg/m^3 on an object with volume (V) is:

Fb = ρ*g*V = 1.225*9.81*V = 12.02*V

- For the similar object with mass (m), the downward weight would be:

W = m*g

- For the object to float the buoyant force (Fb) must be greater than weight of the object:

Fb > W

12.02*V > m*9.81

V / m > 0.816

- The ratio of V / m must be at-least = 0.816.

- Assuming the object is fully immersed in air, then the volume displaced V = ρ_material*V

ρ_material < 1 / 0.816

ρ_material < 1.225 or ( ρ_air )

- So the for an object to float in air its material density must always be less than that of air. That why in balloons lighter gas is used which have density less than that of air like Helium.

4 0

4 years ago