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3 years ago

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A 4.5 kg mass is accelerated at 40 m/s/s what is the force that was applied

1 answer:

cestrela7 [59]3 years ago

8 0

Answer:

180 N

Explanation:

Use Newton's second law, F = ma.

m = 4.5 kg; a = 40 m/s/s

F = (4.5)(40) = 180 N

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Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same tim

blagie [28]

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

$x=x_{0}+V t \\ x=\frac{1}{2}+V t$

The distance between the police car and the intersection is,

$y=y_{0}+V t$

$y=\frac{1}{2}-40 t$

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

$z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })$

$z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)$

The rate of change is,

$2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)$

$2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots$

Now finding $z$ when $t=0,$ from (1) we have

$z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}$

$z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071$

The officer's radar gun indicates 25 mph pointed at the other car then, $\frac{d z}{d t}=25$ when $t=0,$ from

From (2) we get

$2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)$

$2(0.7071)(25)=V+2 V^{2}(0)-40$

$35.36=V-40$

$V=35.36+40=75.36$

Hence the speed of the car is $75.36 mph$

7 0

2 years ago

In the diagram, q1, q2, and q3 are in a straight line.

schepotkina [342]

Answer:

did anyone figure it out

Explanation:

please help

3 0

2 years ago

What is the flow rate of water in a pipe flowing full with an area of 0.3 m2 and velocity of 2.5 m/s?

sashaice [31]

Answer:

0.75 m³/s

Explanation:

Applying,

Q = vA.................... Equation 1

Where Q = flow rate of the water, v = velocity of the water, A = area of the pipe.

From the question,

Given: v = 2.5 m/s, A = 0.3 m²

Substitute these values into equation 1

Q = 2.5(0.3)

Q = 0.75 m³/s

Hence the flow rate of water in the pipe is 0.75 m³/s

4 0

2 years ago

A low C (f=65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and

WITCHER [35]

Answer:

T = 676 N

Explanation:

Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g $/m^{2}$ = 0.005 kg

A stationary wave that is set up in the string has a frequency of;

f = $\frac{1}{2L}$ $\sqrt{\frac{T}{M} }$

⇒ T = 4 $L^{2}$ $f^{2}$ M

Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.

But M = L × ρ = (2 × 0.005) = 0.01 kg/m

T = 4 × $2^{2}$ × $65^{2}$ × 0.01

= 4 × 4 ×4225 × 0.01

= 676 N

Tension of the wire is 676 N.

4 0

3 years ago

A body of volume 2000 cm3 has a mass 4 kg. Find the density of the body. Will the body sink or float in water, if the density of

Brums [2.3K]

Answer:

$2 g/cm^3$ , it will sink

Explanation:

The density of an object is given by

$d=\frac{m}{V}$

where

m is the mass of the object

V is its volume

For the body in the problem, we have

m = 4 kg = 4000 g

$V=2000 cm^3$

Therefore, its density is

$d=\frac{4000}{2000}=2 g/cm^3$

And the object will sink in water, because its density is larger than that of water, which is $1 g/cm^3$ . (an object sinks when its density is larger than that of water, otherwise it floats).

6 0

3 years ago