3 years ago

A body of mass m moves along y such that at time t its position is y(t)= at^2 – bt + c, where a, b, c are constant.

Physics

1 answer:

Akimi4 [234]3 years ago

3 0

Answer:

what do u need help with

Explanation:

A body of mass m moves along X-axis such that its position co-ordiante at any instant t is x = `at^(4)

You might be interested in

A differnece in electric potential is required for an electric charge to flow through a wire.

cupoosta [38]

Answer:

TRUE

Explanation:

6 0

3 years ago

Read 2 more answers

At an accident scene on a level road, investigators measure a car's skid mark to be 93 m long. It was a rainy day and the coeffi

USPshnik [31]

Answer:

u = 25 m/s

Explanation:

given,

length of skid = 93 m

coefficient of friction = 0.35

final velocity = 0 m/s

initial velocity = ?

force here is friction f = μ mg

F = ma

now com paring

-μ mg = m a

a = - μ g

a = - 0.35 x 9.8

a = -3.43 m/s²

we know,

v² = u² + 2 a s

0 = u² - 2 x 3.43 x 93

u² = 637.98

u = 25.26 m/s

u = 25 m/s (two significant figure)

6 0

3 years ago

"One step in the manufacture of silicon wafers used in the microelectronics industry is the melt crystallization of silicon into

nevsk [136]

Answer:

fgfgfdgdg

Explanation:

gfdgfgfgd

7 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 = $\frac{h_{liquid} }{ h_{body} }$

8 0

2 years ago

a watermelon is balanced by a dog, a pumpkin, a flowerpot, and a baseball as shown below. Is the weight of the watermelon equal

alukav5142 [94]

Answer:

pumpkin

Explanation:

watermelon and pumpkins are close to shape and size

6 0

3 years ago

Read 2 more answers