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2 years ago

5

How is the law of acceleration applied in accomplishing a task in relation to a person's ability and the effort exerted?

1 answer:

ivann1987 [24]2 years ago

8 0

Newton's second law of motion states that the force is equal to the product of the mass and acceleration. It can mathematically be expressed as,
F = m x a
When this equation is transformed to solve for acceleration, this becomes,
a = F / m
The equation suggests that for an object to move, force has to be exerted.

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(SCIENCE) 22 POINTS AND CROWN ASAP

gayaneshka [121]

Answer:

A or D

Explanation:

Net force includes addition or subtraction.

5 0

2 years ago

A 0.35-kgkg cord is stretched between two supports, 7.4 mm apart. When one support is struck by a hammer, a transverse wave trav

Agata [3.3K]

Answer:

The tension in the string is $T = 1.49*10^{-6}N$ .

Explanation:

For a string with tension $T$ and linear density $\mu_d$ carrying a transverse wave at speed $v$ it is true that

$v = \sqrt{\dfrac{T}{\mu_d} }$

solving for $T$ we get:

$T = \dfrac{v^2}{\mu_d}.$

Now, the transverse wave covers the distance of 7.4mm in 0.88s, which means it's speed is

$v =\dfrac{7.4*10^{-3}m}{0.88s} \\\\v = 8.4*10^{-3}s$

And it's linear density (mass per unit length) is

$\mu_d = \dfrac{0.35kg}{7.4*10^{-3}m} \\\\\mu_d = 47.3kg/m$

Therefore, the tension in the cord is

$T = \dfrac{(8.4*10^{-3}m/s^2)^2}{47.3kg/m}.$

$\boxed{T = 1.5*10^{-6}N}$

or in micro newtons

$T =1.5\mu N$

4 0

2 years ago

A 50kg boy runs at a speed of 10.0m/s and jumpsonto a cart

andriy [413]

Answer:

the mass of the cart is 150 kg

Explanation:

given,

mass of boy(m) = 50 kg

speed of boy (v)= 10 m/s

initial velocity of cart (u) = 0

final velocity of cart(V) = 2.5 m/s

mass of the cart(M) = ?

m v + M u = (m + M ) V......................(1)

50× 10 + 0 = (50 + M ) 2.5

M = $\dfrac{500}{2.5} - 50$

M = 150 Kg

hence, the mass of the cart is 150 kg

4 0

2 years ago

Read 2 more answers

An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in

Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

$T {V}^{\gamma - 1} = \text{constant}$

$\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}$

Given:-

${T}_{1} = 275 \; K$

${T}_{2} = T \left( \text{say} \right)$

${V}_{1} = V$

${V}_{2} = 6V$

$\gamma = \cfrac{5}{3} \;$ (the gas is monoatomic)

$\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}$

$\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}$

T = 275 $\times$ 0.30

T = 83 K.

3 0

3 years ago

Mathew throws a ball straight up into the air. It rises for a period of time and then begins to drop. At which points in the bal

Cloud [144]

When It begins to drop because that when gravity will have its strongest pull on the object.

8 0

2 years ago