Ask question

Ask question

All categories

1 year ago

14

One object has twice as much mass as another object. The first object also has twice as much a velocity. b gravitational acceler

ation. c inertia. d all of the above

1 answer:

Gemiola [76]1 year ago

6 0

It would be c. As it can’t accelerate faster thus not having a faster velocity so it’s inertia

You might be interested in

What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56×104v/m and a magnetic fiel

sashaice [31]

1) $3.38\cdot 10^6 m/s$

When both the electric field and the magnetic field are acting on the electron normal to the beam and normal to each other, the electric force and the magnetic force on the electron have opposite directions: in order to produce no deflection on the electron beam, the two forces must be equal in magnitude

$F_E = F_B\\qE = qvB$

where

q is the electron charge

E is the magnitude of the electric field

v is the electron speed

B is the magnitude of the magnetic field

Solving the formula for v, we find

$v=\frac{E}{B}=\frac{1.56\cdot 10^4 V/m}{4.62\cdot 10^{-3} T}=3.38\cdot 10^6 m/s$

2) 4.1 mm

When the electric field is removed, only the magnetic force acts on the electron, providing the centripetal force that keeps the electron in a circular path:

$qvB=m\frac{v^2}{r}$

where m is the mass of the electron and r is the radius of the trajectory. Solving the formula for r, we find

$r=\frac{mv}{qB}=\frac{(9.1 \cdot 10^{-31} kg)(3.38\cdot 10^6 m/s)}{(1.6\cdot 10^{-19} C)(4.62\cdot 10^{-3}T)}=4.2\cdot 10^{-3} m=4.1 mm$

3) $7.6\cdot 10^{-9}s$

The speed of the electron in the circular trajectory is equal to the ratio between the circumference of the orbit, $2 \pi r$ , and the period, T:

$v=\frac{2\pi r}{T}$

Solving the equation for T and using the results found in 1) and 2), we find the period of the orbit:

$T=\frac{2\pi r}{v}=\frac{2\pi (4.1\cdot 10^{-3} m)}{3.38\cdot 10^6 m/s}=7.6\cdot 10^{-9}s$

7 0

2 years ago

Un alambre de 2m de longitud soporta una carga de 5400 N. El diámetro de la sección transversal es de 2 mm y el módulo de Young

tester [92]

Hey there, hope you’re having a positive day. Can you try translate this into English please, unfortunately I do not understand your question as I don’t understand your language. No hate towards you happy Thanksgiving

3 0

1 year ago

While attempting a landing on the moon, astronauts had to change their landing site and land at a spot that was 4 kilometers awa

lesya [120]

The closest answer would be C.
The choices given do not give the exact value.

To answer this, you just need to remember the main formula:

$d = Vit + \frac{1}{2}gt^{2}$

Where:
d = distance/displacement
g = acceleration due to gravity
t = time in flight
Vi = initial velocity.

With this formula, you derive all the formulas you need to look for certain components. You need to keep in mind of the following:

---if you are looking for a vertical component(y), you need to use values of vertical motion.

$dy = Viyt + \frac{1}{2}gt^{2}$

*Viy is always 0m/s at the beginning of a free-fall.

$dy = (0m/s)t + \frac{1}{2}gt^{2}$

$dy = \frac{1}{2}gt^{2}$

---If you are looking for a horizontal component(x), you need to use values of horizontal motion.

$dx = Vixt + \frac{1}{2}gt^{2}$

*g is always 0m/s² when taking horizontal motion into account.

$dx = Vixt + \frac{1}{2}(0m/s^{2})t^{2}$

$dx = Vixt$
--- time is the only value that is both vertical and horizontal.

Okay, let's get back to solving your problem. Let's see what your given is first:
dy = 137m (as long as it refers to height, it is vertical distance)
dx = 4km (the word, far or away usually indicates horizontal distance)
g = 1.63m/s²

The question is how fast was it going horizontally and we can derive it from our equation:

$dx = Vixt$

We use this because x means horizontal. But notice that we do not have time yet. So how are we going to solve this 2 variables missing? The key is that time is a horizontal and vertical component. Whatever time it took moving horizontally, it is the same vertically as well. So we use the vertical formula to derive time:

$dy = \frac{1}{2}gt^{2}$
$\frac{2dy}{g}=t^{2}$
$\sqrt{\frac{2dy}{g}} = \sqrt{t^{2}}$
$\sqrt{\frac{2dy}{g}} = t$

Now plug in what you know and solve for what you don't know:

$\sqrt{\frac{2(137m)}{1.63m/s^{2}}} = t$
$\sqrt{\frac{274m}{1.63m/s^{2}}} = t$
$\sqrt{168.098s^{2}}=t$
$12.965s = t$

The total time in flight is 12.965s.
Let's round it off to 13s.

Now that we know that, we can use this in the horizontal formula:
$dx = Vixt$
$4km = Vix(13s)$

Hold up! Look at the unit of the horizontal distance. It is in km but all our units are expressed in m so we need to convert that first.

1km = 1,000m
4km = 4,000m

Our new horizontal distance is 4,000m.

Okay, let's wrap this up by solving for what is asked for, using all the derived values.
$dx = Vixt$
$4,000m = Vix(13s)$
$\frac{4,000m}{13s} = Vix$
$308m/s= Vix$

The horizontal velocity is 308m/s.

Such a long explanation I know, but hopefully, you learned from it.

6 0

2 years ago

A projectile is fired from ground level with a speed of 150 m/s at an angle 30.° above the horizontal on an airless planet where

77julia77 [94]

Answer:

129.9 m/s = 130 m/s to 2 s.f

Explanation:

For projectile motion, the initial velocity combined with the angle of launch is used to obtain the initial horizontal and vertical components of the velocity.

u = initial velocity of the projectile = 150 m/s

uₓ = u cos θ = 150 cos 30° = 129.9 m/s

uᵧ = u sin θ = 150 sin 30° = 75.0 m/s

In the motion of a projectile, the motion can literally be separated into vertical and horizontal components.

The vertical component has to do with the acceleration due to gravity (acting downwards) on the projectile and the vertical component of the velocity (which changes all through the motion of the projectile because of the force of gravity manifested in the form of acceleration due to gravity).

But there is no force acting in the horizontal direction, hence, no acceleration in the horizontal direction for projectile motion. This directly translates to a constant velocity in the horizontal direction all through the flight of the projectile.

Hence, the horizontal component of the velocity of the projectile at t = 4 s is the same as the horizontal component of the initial velocity.

Horizontal component of the Velocity at t = 4s is equal to uₓ = u cos θ = 150 cos 30° = 129.9 m/s = 130 m/s

6 0

1 year ago

Read 2 more answers

Assuming both graduated cylinders are holding water at room temperature, which cylinder has more thermal energy?

natta225 [31]

Answer:

The correct option is;

The graduate cylinder with more water has more thermal energy because it is holding more water molecules

Explanation:

Given that the thermal energy of the system is the energy possessed by the system by virtue of the increased motion of the particles by virtue of a transfer of heat, when the content of the system is heated

The thermal energy, Q is given by the following equation;

Q = Mass, m × The specific heat capacity, C × The change in temperature, ΔT

Given that the graduated cylinder with more water has more mass and therefore, more water molecules, than the cylinder with less water, the cylinder with more water has more thermal energy.

3 0

2 years ago