Help please <br><br><br> -spaceholder this is-

You could use newton’s second law to calculate the force applied to an object if you knew the objects mass and its _____.

olga2289 [7]

Answer:

You could use newton’s second law to calculate the force applied to an object if you knew the objects mass and its <u>acceleration.</u>

Explanation:

By, Newtons second law, the force applied on an object directly varies with the acceleration caused and the mass of the object.

This is given by :

$F=m\ a$

Where $F$ represents force applied on the object , $m$ represents mass of the object and $a$ represents the acceleration.

In order to calculate force applied on object we require the mass of the object and its acceleration. The force can be calculated by finding the product of mass and acceleration of the object.

4 0

3 years ago

A point charge of 5.7 µc moves at 4.5 × 105 m/s in a magnetic field that has a field strength of 3.2 mt, as shown in the diagram

Sauron [17]

The magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

<h3>What is magnetic force?</h3>

Magnetic force, attraction or repulsion that arises between electrically charged particles because of their motion.The magnitude of the magnetic force acting on the charge is given by:

$\rm F=qvBsin\theta$

where

The magnitude of the charge $q=5.7\ \mu C =5.7\times 10^{-6} \ C$

The velocity of the charge $v=4.5\times 10^5\ \frac{m}{s}$

The magnitude of the magnetic field $3.2\ mT=0.0032\ T$

The angle between the directions of v and B $\theta =90^o-37^o=53^o$

By substituting the values we will get:

$F=(5.7\times 10^}-6})(4.5\times 10^5)(0.0032)(Sin53^o)$

$F=6.6\times 10^{-3}\ N$

Hence the magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

To know more about Magnetic force follow

brainly.com/question/14411049

3 0

1 year ago

You are on roller blades on top of a small hill. Your potential energy is equal to 1,000.0 joules. The last time

irinina [24]

It is -59 i think or just ask someone esle for help

8 0

3 years ago

An ice cube at 0c was dropped into 30.0 g of water in a cup at 45.0c. at the instant that all of the ice was melted, the tempera

Ede4ka [16]

The amount of heat given by the water to the block of ice can be calculated by using
$Q=m_w C_{sw} \Delta T_w$
where
$m_w = 30 g$ is the mass of the water
$C_{sw}=4.18 J/(g ^{\circ}C)$ is the specific heat capacity of water
$\Delta T_w = 45.0^{\circ}-19.5^{\circ}C = 20.5^{\circ}C$ is the variation of temperature of the water.

Using these numbers, we find
$Q=(30 g)(4.18 J/(g^{\circ}C))(20.5^{\circ}C)=2571 J$

This is the amount of heat released by the water, but this is exactly equal to the amount of heat absorbed by the ice, used to melt it into water according to the formula:
$Q = m_i L_f$
where $m_i$ is the mass of the ice while $L_f =334 J/g$ is the specific latent heat of fusion of the ice.
Re-arranging this formula and using the heat Q that we found previously, we can calculate the mass of the ice:
$m_i = \frac{Q}{L_f}= \frac{2571 J}{334 J/g} =7.7 g$

3 0

2 years ago

A drag racing vehicle travels from zero to a hundred miles per hour and 5 seconds north what is acceleration

avanturin [10]

Acceleration(a) is the change in velocity(Δv) over the change in time(Δt). so just divide your velocity and time.

6 0

2 years ago

Help please -spaceholder this is-