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Oksana_A [137]
3 years ago
12

A 150kg person is pushed from the behind and accelerates at 5 m/s2. What is the net force applied to the person? ANSWER FOR BRAI

NLIEST!
Physics
1 answer:
kogti [31]3 years ago
6 0

Explanation:

Force = Mass × Acceleration

Mass = 150kg

Acceleration = 5 m/s^2

Force = 150 × 5 = 750N

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The car's initial speed was 15 m / s and the distance the car travels before it comes to a complete stop after the driver applie
pentagon [3]

Initial speed of the car (u) = 15 m/s

Final speed of the car (v) = 0 m/s (Car comes to a complete stop after driver applies the brake)

Distance travelled by the car before it comes to halt (s) = 63 m

By using equation of motion, we get:

\bf \longrightarrow  {v}^{2}  =  {u}^{2}  + 2as \\  \\ \rm \longrightarrow  {0}^{2}  =  {15}^{2}  + 2 \times a \times 63 \\  \\ \rm \longrightarrow 0 = 225 + 126a \\  \\ \rm \longrightarrow 126a =  - 225 \\  \\ \rm \longrightarrow a =  -  \dfrac{225}{126}  \\  \\ \rm \longrightarrow a =  - 1.78 \: m {s}^{ - 2}

\therefore Acceleration of the car (a) = -1.78 m/s²

Magnitude of the car's acceleration (|a|) = 1.78 m/s²

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3 years ago
A student is planning to make changes to a circuit Which change will
garik1379 [7]
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3 years ago
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What has research determined about the orbit of an electron around a nucleus?
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4 0
3 years ago
A 40 lb box is suspended from two ropes which each make an angle of 45 degrees with the vertical. What is the tension (in pounds
givi [52]

Answer:

The tension on each rope is 28.28 lb

Explanation:

Given that,

Weight = 40 lb

Angle = 45°

We need to calculate the tension on each rope

Considering the symmetry of the system, the tension of each rope must be same

So,

T\sin\theta+T\sin\theta=mg

Put the value into the formula

2T\sin\theta=mg

T\times\sin45=\dfrac{40}{2}

T=\dfrac{40\sqrt{2}}{2}

T=28.28\ lb

Hence, The tension on each rope is 28.28 lb.

3 0
3 years ago
The opening to a cave is a tall, 30-cm-wide crack. A bat that is preparing to leave the cave emits a 30 kHz ultrasonic chirp. Ho
sweet-ann [11.9K]

Answer:

Width of sound beam is 7.557 m

Explanation:

First we will calculate the wave length from given data:

λ=v/f

Were:

v is the speed

f is the frequency

\lambda=\frac{340}{30*10^3}\\ \lambda=0.01133 m

We considered the opening long and narrow, Using single slit diffraction formula:

mλ=dsinΘ

where:

d is the crack width

m is the order

Θ is angle

Considering m=1, The angle between first minimum from center of beam is:

\theta=sin^{-1}(\frac{m\lambda}{d})\\\theta=sin^{-1}(\frac{1*0.01133}{30*10^{-2}})\\ \theta=2.164^o

The width of beam is:

tanΘ=y/L

tan\theta=\frac{w/2}{L}\\ w=2L\ tan\theta\\w=2*100*tan 2.164\\w=7.557 m

Width=7.557 m

5 0
3 years ago
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