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3 years ago

I need help!

Physics

1 answer:

nata0808 [166]3 years ago

7 0

Answer:

The acceleration of the car with the mass of 1200kg and a force of 11 × 10³N is 9.17m/s²

Explanation:

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It takes 6 sec for a stone to fall the the bottom of a mineshaft. calculate the

Stels [109]

Ok so this is simple projectile motion problem.

if we have an object falling in free fall it is subject to gravity of -9.80m/s^2

so it says it takes 6 sec to fall and we know initial velocity was zero so we know that h=vt+1/2gt^2 so we get h=0+1/2*9.80*6^2 = 176.4m

so solving for final speed we get KE=PE = 1/2mv^2=mgh = 1/2v^2=gh so
v=sqrt(2*g*h) = sqrt(2*9.8*176.4m) = 58.8m/s final speed when it hits the ground

hope this helps you! Thanks!!

8 0

3 years ago

A 20 cm square frame that can rotate about the 00' axis is placed in a homogeneous magnetic field with 0.5T induction directed v

IRISSAK [1]

The solution to this task is f00king Dis.

6 0

2 years ago

How long (in seconds) does it take a car accelerating at 3.1 m/s2 to go from rest<br> to 51 m/s?

Nesterboy [21]

Answer:

t = 16.5s

Explanation:

Given parameters:

Acceleration = 3.1m/s²

Initial velocity = 0m/s

Final velocity = 51m/s

Unknown:

Time taken = ?

Solution:

To solve this problem we need to reiterate that acceleration is the rate of change of velocity with time.

So;

Acceleration = $\frac{v - u }{t}$

v is the final velocity

u is the initial velocity

t is the time taken

So;

3.1 = $\frac{51 - 0}{t}$

3.1t = 51

t = 16.5s

8 0

3 years ago

a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f

34kurt

Answer:

$\frac{h_{liquid} }{ h_{body} }$ = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

B = ρ_liquid g V_liquid

let's write the translational equilibrium condition

B - W = 0

let's use the definition of density

ρ_body = m / V_body

m = ρ_body V_body

W = ρ_body V_body g

we substitute

ρ_liquid g V_liquid = ρ_body g V_body

$\frac{\rho_{body} }{\rho_{liquid} } } = \frac{V_{liquid} }{V_{body} } }$

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

V = A h_bogy

Thus

$\frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }$

we substitute

5/9 = $\frac{h_{liquid} }{ h_{body} }$

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2 years ago

Soft ball,kick ball,wiffle ball??

Anvisha [2.4K]

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2 years ago

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