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marysya [2.9K]
4 years ago
9

A Tesla Roadster car accelerates from rest at a rate of 7.1m/s for a time of 3.9s

Physics
1 answer:
liberstina [14]4 years ago
5 0

The distance covered is 54.0 m

Explanation:

Since the motion of the car is a uniformly accelerated motion, we can use the following suvat equation:

s=ut+\frac{1}{2}at^2

where

u is the initial velocity

t is the time

a is the acceleration

s is the distance covered

For the car in this problem, we have

u = 0 (it starts from rest)

a=7.1 m/s^2 is the acceleration

t = 3.9 s is the time

Substituting, we find s:

s=0+\frac{1}{2}(7.1)(3.9)^2=54.0 m

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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Write a explanation explaining why red is a beautiful color is not a hypothesis
PIT_PIT [208]

Answer:

Please find the explanation below

Explanation:

A hypothesis in science is a testable explanation that is yet to be tested via experimentation. It is a predictive statement or suggested solution to an observation. A hypothesis aims at finding a possible explanation/answer to a question, which is subject to testing. One important aspect of formulating a hypothesis is that it tends to connect the independent variable with the dependent/measurable variable.

The statement "RED IS A BEAUTIFUL COLOR" cannot be considered a hypothesis because it does not aim to answer a question that can undergo experimental testing. This statement can not be measured via experimentation. The statement is not a possible answer to a question but rather a personal opinion about something.

5 0
3 years ago
A block of wood 3.0 cm on each side has a mass of 27g. what is the density of this block?
schepotkina [342]
The block of wood is 3cm on each side so it is a cube. The volume of a cube is given by s^3. So the volume of this block is 3cm x 3cm x3 cm = 27 cm^3. density = mass/volume =27 g / 27 cm^3 = 1 g/cm^3
8 0
3 years ago
1. A step-up transformer increases 15.7V to 110V. What is the current in the secondary as compared to the primary? Assume 100 pe
Serjik [45]

1. I_2 = 0.14 I_1

Explanation:

We have:

V_1 = 15.7 V voltage in the primary coil

V_2 = 110 V voltage in the secondary coil

The efficiency of the transformer is 100%: this means that the power in the primary coil and in the secondary coil are equal

P_1 = P_2\\V_1 I_1 = V_2 I_2

where I1 and I2 are the currents in the two coils. Re-arranging the equation, we find

\frac{I_2}{I_1}=\frac{V_1}{V_2}=\frac{15.7 V}{110 V}=0.14

which means that the current in the secondary coil is 14% of the value of the current in the primary coil.

2. 5.7 V

We can solve the problem by using the transformer equation:

\frac{N_p}{N_s}=\frac{V_p}{V_s}

where:

Np = 400 is the number of turns in the primary coil

Ns = 19 is the number of turns in the secondary coil

Vp = 120 V is the voltage in the primary coil

Vs = ? is the voltage in the secondary coil

Re-arranging the formula and substituting the numbers, we find:

V_s = V_p \frac{N_s}{N_p}=(120 V)\frac{19}{400}=5.7 V

5 0
3 years ago
A 1300 kg steel beam is supported by two ropes. (Figure
Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

  • the net horizontal force acting on the beam is

R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

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R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

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2 years ago
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makkiz [27]

Answer:

0 N

Explanation:

suppose, you push a box with 5 N, and another person pushes the box on the opposite side of the box with 5 N, the net force (resultant ) is 0 N, the box will not move if it wasn't moving

hope this helps

8 0
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