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3 years ago

7

What force is required to accelerate a body with a mass of 15 kilograms at a rate of 8 m/s²?

1 answer:

lawyer [7]3 years ago

4 0

I really think it could be a 23 kg

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Ifa truck starts from rest and it has acceleration of 4 m/s for 5 second

yan [13]

Answer:

Distance 50m

final velocity 20ms^-1

$s = ut + \frac{1}{2} a {t}^{2} \\ = 0 \times t + \frac{1}{2} \times 4 \times {5}^{2} = 50m$

$v = u + at \\ v = 0 + 4 \times 5 \\ v = 20ms^{ - 1}$

6 0

2 years ago

Acid rain is the direct result of acid released by burning fossil fuels. T/F

disa [49]

Answer : False.

Explanation : Acid rain is the indirect result of burning fossil fuels. As the incomplete combustion of fossil fuels releases sulfur dioxide and nitrous dioxide in the air. When it rains it gets mixed with water and forms nitric acid and sulfuric acid which causes acid rain. Acid rain is harmful.

7 0

3 years ago

Read 2 more answers

The greater the mass of an object, the greater its force due to gravity

Serga [27]

What is the question?

6 0

3 years ago

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

3 years ago

What does the diameter in meters need to be of a round parachute that the designer has determined needs to have an area of 500 s

Marrrta [24]

Area of a circle is
A= pi r^2
so
500m^2 = 3.14 r2
500/pi = r ^2
152.1549...=r^2
square root both sides
r=12.61566...
d=2r
d=25.2
to 3 sig fig

4 0

3 years ago