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2 years ago

Hii! help asap. i’ll give brainliest thanks!

Physics

1 answer:

telo118 [61]2 years ago

7 0

Its acceleration will change

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A 1kg box is pushed on a flat surface that is 250m long. The box is initially at rest and then pushed with a constant Net force

sasho [114]

Answer:

C) 50 m/s

Explanation:

With the given information we can calculate the acceleration using the force and mass of the box.

Newton's 2nd Law: F = ma

5 N = 1 kg * a
a = 5 m/s²

List out known variables:

v₀ = 0 m/s
a = 5 m/s²
v = ?
Δx = 250 m

Looking at the constant acceleration kinematic equations, we see that this one contains all four variables:

v² = v₀² + 2aΔx

Substitute known values into the equation and solve for v.

v² = (0)² + 2(5)(250)
v² = 2500
v = 50 m/s

The final velocity of the box is C) 50 m/s.

7 0

3 years ago

Plssss help Do you guys have any other more interesting, funny or creative ideas of incline problems than something sliding down

VashaNatasha [74]

U can always just do the classic roller coaster going up an incline and create some sort of story from that.

8 0

3 years ago

What is the specific heat of an object if it takes 1200 J to raise the temperature of a 20 kg object by 6 degrees C?

ch4aika [34]

Answer:

$c=10\ J/kg^{\circ} C$

Explanation:

Given that,

Heat required, Q = 1200 J

Mass of the object, m = 20 kg

The increase in temperature, $\Delta T=6^{\circ} C$

We need to find the specific heat of the object. The heat required to raise the temperature is given by :

$Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{1200}{20\times 6}\\\\c=10\ J/kg^{\circ} C$

So, the specific heat of the object is $10\ J/kg^{\circ} C$ .

5 0

2 years ago

A 180 lb crate is on the ground, and a strong rope is attached. You need to move it across the basement floor, which has a coeff

jekas [21]

Answer:

$F = 505.13 N$

Yes it is better to pull the rope rather than push it

Explanation:

Let the force is applied at an angle of 60 degree

so we will have net vertical force on the crate is given as

$F_n + Fsin60 = mg$

here we know

$m = 180 lb$

$m = 81.65 kg$

$F_n = 81.65(9.81) - Fsin60$

$F_n = 801 - 0.866 F$

now friction force on the crate is given as

$F_x = \mu F_n$

$Fcos60 = 0.7(801 - 0.866 F)$

$0.5F + 0.61F = 560.7$

$F = 505.13 N$

Yes it is better to pull the rope rather than push it

6 0

3 years ago

If two solutions have unequal concentrations of a solute, the solution with the lower concentration is called

marta [7]

Obviouskly lower concentrstion

6 0

2 years ago