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2 years ago

7

Use the following statements to determine if the disjunction is true or false. p: 7 is an odd number q: 3 is the square root of

9 is p q true or false? true false

1 answer:

kotykmax [81]2 years ago

3 0

P and q are both true

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As altitude increases in the troposphere and stratosphere, the air temperature does what?

natima [27]

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1

Explanation:

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2 years ago

You are riding in a vehicle with a coffee cup on the dashboard while traveling 30 km/h. The vehicle makes a sudden stop to avoid

wariber [46]

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It would move forward.

Explanation:

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3 years ago

A team of dogs accelerates a 290kg dogsled from 0 to 6.0m/s in 3.0 s. Assume that the acceleration is constant.Part AWhat is the

Mademuasel [1]

Answer:

(a) $a=2m/sec^2$

(b) 5220 j

(c) 1740 watt

(d) 3446.66 watt

Explanation:

We have given mass m = 290 kg

Initial velocity u = 0 m/sec

Final velocity v = 6 m/sec

Time t = 3 sec

From first equation of motion

v = u+at

So $a=\frac{v-u}{t}=\frac{6-0}{3}=2m/sec^2$

(a) We know that force is given by

F = ma

So force will be $F=290\times 2=580N$

(b) From second equation of motion we know that

$s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 2\times 3^2=9m$

We know that work done is given by

W = F s = 580×9 =5220 j

(c) Time is given as t = 3 sec

We know that power is given as

$P=\frac{W}{t}=\frac{5220}{3}=1740Watt$

(d) Time t = 1.5 sec

So $P=\frac{W}{t}=\frac{5220}{1.5}=3466.66Watt$

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3 years ago

When you are riding in a car then the driver suddenly steps on the gas pedal, why do you feel a jolt and push back?

inessss [21]

Answer:

due to acceleration lol

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2 years ago

Read 2 more answers

A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is

pochemuha

Answer:

2856.96 J

0

0

$\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

6.78822 m/s

Explanation:

$v_i$ = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J$

At height y = 0 the potential energy is 0 as

$P=mgy\\\Rightarrow P=mg0=0$

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

$K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

Half of maximum height

$\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}$

$h_i=\frac{v_i^2}{2g}$

$v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s$

The velocity of the athlete at half the maximum height is 6.78822 m/s

8 0

2 years ago