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Sedaia [141]
4 years ago
12

A net force of 125 n is applied to a certain object. as a result, the object accelerates with an acceleration of 24.0 m/s2. the

mass of the object is
Physics
2 answers:
pantera1 [17]4 years ago
8 0
Newton's second law states that Fnet = ma, where Fnet is the net force applied, m is the mass of the object, and a is the object's acceleration. You have the values for Fnet and a, so you simply use this equation to solve for m, mass.
Darina [25.2K]4 years ago
4 0

Answer:

m=5.2kg

Explanation:

A net force of 125 n is applied to a certain object. as a result, the object accelerates with an acceleration of 24.0 m/s2. the mass of the object is

from newton's second law of motion , which states that

the rate change in momentum of an object is directly proportional to the force applied

definition of terms  you should know.

a force is tat which tends to cane a body state of rest or uniform motion in a strait line

mass is the quantity of matter in a body

acceleration is change in velocity per time

therefore,

f∝m(V-U)/t

f=km(V-U)/t

k=1

f=m(V-U)/t

(V-U)/t=a

f=ma

f=125N

a=24m/s^2

m=5.2kg

the mass of the object

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The force diagram represents a girl pulling a sled with a mass of 6.0 kg to the left with a force of 10.0 N at a 30.0 degree ang
STatiana [176]

the correct answers are 54N and -1,2m/s^2

6 0
3 years ago
Read 2 more answers
An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an i
Lina20 [59]

Answer:

1317.4 m

Explanation:

We are given that

Angle=\alpha=40.8^{\circ}

Initial speed =v_0=346m/s

We have to find the horizontal distance covered  by the shell after 5.03 s.

Horizontal component of initial speed=v_{ox}=v_0cos\theta=346cos40.8=261.9m/s

Vertical component of initial speed=v_{oy}=346sin40.8=226.1m/s

Time=t=5.03 s

Horizontal distance =Horizontal\;velocity\times time

Using the  formula

Horizontal distance=261.9\times 5.03

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m

8 0
3 years ago
Two lasers, one red (with wavelength 633.0 nmnm) and the other green (with wavelength 532.0 nmnm), are mounted behind a 0.150-mm
Ratling [72]

(a) The screen  is 3.20m from the split.

(b) The closest minima for green, distance Δy = 0.45 cm.

When a wave hits a barrier or opening, numerous events are referred to as diffraction. It is described as the interference or bending of waves via an aperture or around the corners of an obstruction into the area that forms the geometric shadow of the obstruction or aperture.

(a)Equation of minima = sinθ  = mλ/α

Given, m = 3, λ = 6.33X10⁻⁷, α = 0.00015

Putting the values in formula to get θ.

  θ = sin⁻¹ ( \frac{3 X 6.33X10^{-7} }{0.00015} ) = 0.01266 rad

triangle need to be drawn to find relationship between θ, y$ and L

tan(θ) = y/L  where; y = 4.05 cm

L = y/tan(θ) = 3.20

Hence, the screen is 3.20m from the split.

(b) Find the closest minima for green

minima equation is sinθ  = mλ/α where, m = 4 (minima with smallest distance)

sinθ  = 4λ/α

θ = sin⁻¹ (\frac{4X6.33X10^{-7} }{0.00015}) = 0.01688 rad

Calculate L using

tanθ = y/L

  L = 4.5 cm

From equation subtract y₃ from y:

                 4.50 cm - 4.05 cm = 0.45 cm

Hence, distance Δy = 0.45 cm.

Learn more about the Diffraction with the help of the given link:

brainly.com/question/12290582

#SPJ4

I understand that the question you are looking for is "Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the ot

Question

Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the other side of the slit is a white screen. When the red laser is turned on, it creates a diffraction pattern on the screen.

a. The distance y3,red from the center of the pattern to the location of the third diffraction minimum of the red laser is 4.05 cm. How far L is the screen from the slit? Express this distance L in meters to three significant figures.

b. With both lasers turned on, the screen shows two overlapping diffraction patterns. The central maxima of the two patterns are at the same position. What is the distance Δy between the third minimum in the diffraction pattern of the red laser (from Part A) and the nearest minimum in the diffraction pattern of the green laser?

5 0
2 years ago
Plz di all plz i will give brainest and thanks to best answer do it right
Inga [223]

Answer:

area 4

Explanation:

area 4 has low pressure

7 0
3 years ago
A firecracker breaks up into two pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.0 m/
larisa [96]

Answer:

Total momentum, p = 21.24 kg-m/s

Explanation:

Given that,

Mass of first piece, m_1=200\ g= 0.2\ kg

Mass of the second piece, m_2=300\ g= 0.3\ kg

Speed of the first piece, v_1=82\ m/s (along x axis)

Speed of the second piece, v_2=45\ m/s (along y axis)

To find,

The total momentum of the two pieces.

Solve,

The total momentum of two pieces is equal to the sum of momentum along x axis and along y axis.

p_x=m_1v_1

p_x=0.2\ kg\times 82\ m/s

p_x=16.4\ kg-m/s

p_y=m_2v_2

p_y=0.3\ kg\times 45\ m/s

p_y=13.5\ kg-m/s

The net momentum is given by :

p=\sqrt{p_x^2+p_y^2}

p=\sqrt{16.4^2+13.5^2}

p = 21.24 kg-m/s

Therefore, the total momentum of the two pieces is 21.24 kg-m/s.

4 0
3 years ago
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