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2 years ago

Which vector should be negative?

Physics

1 answer:

pentagon [3]2 years ago

7 0

Answer:

The velocity of a falling object

Explanation:

The positive X axis is towards right and positive Y axis is towards up, so North direction is positive

A vector with less than 1 magnitude is not negative, because its magnitude may be in between 0 and 1 which is positive vector.

Any vector whose magnitude is greater than 1 is never be a negative vector.

The velocity of a falling object is towards bottom, that is towards negative Y axis. So that vector is negative.

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The distance between two corresponding parts of a wave is its

oksano4ka [1.4K]

Answer:

wavelenght

Explanation:

The wavelength is the spatial period of a wave, analogous to the temporal period, it is the distance between two consecutive points with maximum amplitude that are repeated in space . In the waves of the sea, the wavelength is easily observed in the separation between two consecutive ridges.

4 0

3 years ago

Read 2 more answers

In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.

vaieri [72.5K]

Answer:

(A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

Explanation:

Given that,

A nucleus undergoes a nuclear decay.

(A). In alpha decay,

We know that,

When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.

We need to calculate the resulting nucleus

Using given data

$^{227}_{89}Ac\Rightarrow ^{227-4}_{89-2}X$

$^{227}_{89}Ac\Rightarrow ^{223}_{87}Fr$

The resulting nucleus is Fr.

(B). In beta-minus decay,

We know that,

When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.

We need to calculate the resulting nucleus

Using given data

$^{211}_{83}Bi\Rightarrow ^{211}_{83+1}X$

$^{211}_{83}Bi\Rightarrow ^{211}_{84}Po$

The resulting nucleus is Po.

(C). In beta-plus decay,

We know that,

When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.

We need to calculate the resulting nucleus

Using given data

$^{22}_{11}Na\Rightarrow ^{22}_{11-1}X$

$^{22}_{11}Na\Rightarrow ^{22}_{10}Ne$

The resulting nucleus is Ne.

(D). In gamma decay,

We know that,

When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.

We need to calculate the resulting nucleus

Using given data

$^{98}_{43}Tc\Rightarrow ^{98}_{43}Tc$

The resulting nucleus is Tc.

Hence, (A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

4 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s$

$v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s$

Time taken to reach 1180 m is 11.29 seconds

$v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s$

Time the rocket will keep going up after the engines shut off is 13.06 seconds.

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m$

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s$

Time taken by the rocket to fall from maximum height is 20.29 seconds

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

5 0

3 years ago

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km.A)If you and your spacesuit ha

WITCHER [35]

A) 0.189 N

The weight of the person on the asteroid is equal to the gravitational force exerted by the asteroid on the person, at a location on the surface of the asteroid:

$F=\frac{GMm}{R^2}$

where

G is the gravitational constant

8.7×10^13 kg is the mass of the asteroid

m = 130 kg is the mass of the man

R = 2.0 km = 2000 m is the radius of the asteroid

Substituting into the equation, we find

$F=\frac{(6.67\cdot 10^{-11})(8.7\cdot 10^{13} kg)(130 kg)}{(2000 m)^2}0.189 N=$

B) 2.41 m/s

In order to orbit just above the surface of the asteroid (r=R), the centripetal force that keeps the astronaut in orbit must be equal to the gravitational force acting on the astronaut:

$\frac{GMm}{R^2}=\frac{mv^2}{R}$

where

v is the speed of the astronaut

Solving the formula for v, we find the minimum speed at which the astronaut should launch himself and then orbit the asteroid just above the surface:

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2(6.67\cdot 10^{-11})(8.7\cdot 10^{13} kg)}{2000 m}}=2.41 m/s$

3 0

3 years ago

(1.08×10-3)(9.3×10-4)

ira [324]

First you do the first parenthesis, (1.08 x 10 - 3) and you do it in the order of operations! (parenthesis, exponents, multiplication/division, add/subtract) to get 7.8. Then you take the second parenthesis (9.3 x 10 - 4) and do the same thing to get 89! You then times 7.8 by 89 to get 694.2! If it needs more elaboration just ask ^.^

4 0

3 years ago