All categories

3 years ago

You work at a garden store for the summer, and you lift a 14 kg bag of fertilizer with a force of 227 N.

Physics

1 answer:

nika2105 [10]3 years ago

8 0

Answer:

(a) Acceleration of the bag will be a=16.214m/sec^2

(B) Weight of the bag will be 137.2 N

Explanation:

We have given mass of the bag m = 14 kg

Force with which bag is lifted = 227 N

(A) According to newtons law we force is equal to F = ma , here m is mass and a is acceleration

So $227=14\times a$

$a=16.214m/sec^2$

(b) Acceleration due to gravity $g=9.8m/sec^2$

We know that weight is given by W = mg , here m is mass and g is acceleration due to gravity

So weight $W=mg=14\times 9.8=137.2N$

So weight of the bag will be 137.2 N

You might be interested in

⦁ A car going 50 m/s is brought to rest in a distance of 20.0 m as it strikes a pile of dirt. How large an average force is exer

gtnhenbr [62]

Answer:

the average force exerted by seatbelts on the passenger is 5625 N.

Explanation:

Given;

initial velocity of the car, u = 50 m/s

distance traveled by the car, s = 20 m

final velocity of the after coming to rest, v = 0

mass of the passenger, m = 90 kg

Determine the acceleration of the car as it hit the pile of dirt;

v² = u² + 2as

0 = 50² + (2 x 20)a

0 = 2500 + 40a

40a = -2500

a = -2500/40

a = -62.5 m/s²

The deceleration of the car is 62.5 m/s²

The force exerted on the passenger by the backward action of the car is calculated as follows;

F = ma

F = 90 x 62.5

F = 5625 N

Therefore, the average force exerted by seatbelts on the passenger is 5625 N.

8 0

3 years ago

Which term, taken from the celestial sphere, gives its name to a.m. and p.m.?

DaniilM [7]

Options are. Zenith, Great circle, Equinox, or Meridan

3 0

3 years ago

Suppose we take a 1 m long uniform bar and support it at the 33 cm mark. Hanging a 0.15 kg mass on the short end of the beam res

MakcuM [25]

Answer:

The mass of the beam is 0.074 kg

Explanation:

Given;

length of the uniform bar, = 1m = 100 cm

Set up this system with the given mass and support;

0-----------------33cm-----------------------------------100cm

↓ Δ ↓

0.15kg m

Where;

m is mass of the uniform bar

Apply the principle of moment to determine the value of "m"

sum of anticlockwise moment = sum of clockwise moment

0.15kg(33 - 0) = m(100 - 33)

0.15(33) = m(67)

$m = \frac{0.15kg(33 \ cm)}{67 \ cm}\\\\m = 0.074 \ kg$

Therefore, the mass of the beam is 0.074 kg

8 0

3 years ago

A ball is thrown vertically downward at 10 m/s. what is it’s speed 1s and 2s later

soldi70 [24.7K]

Answer:

20 m/s

30 m/s

Explanation:

Given:

v₀ = -10 m/s

a = -9.8 m/s²

When t = 1 s:

v = v₀ + at

v = (-10 m/s) + (-9.8 m/s²) (1 s)

v = -19.8 m/s

When t = 2 s:

v = v₀ + at

v = (-10 m/s) + (-9.8 m/s²) (2 s)

v = -29.6 m/s

Rounded to one significant figures, the speed of the ball at 1 s and 2 s is 20 m/s and 30 m/s, respectively.

3 0

3 years ago

Many years ago, scientists believed that an atom was the smallest unit of matter. Eventually, evidence was discovered that indic

dybincka [34]

Advances in technology used to study and observe atoms lead to the discovery of electrons, protons, nuetrons, and the quarq

7 0

3 years ago