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krok68 [10]
3 years ago
10

In 3 seconds a car moving in a straight line increases its speed from 22.4 m/s to 29.1 m/s while a truck increases its speed fro

m 0 mph to 6.7m/s in the same amount of time. What is the acceleration of the truck in m/s? *
SHOW ALL WORK
Physics
1 answer:
ANTONII [103]3 years ago
6 0
Is there a picture?
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The magnetic field in the region between the poles of an electromagnet is uniform at any time, (1 point) but is increasing at th
olga nikolaevna [1]

Answer:

B)

The magnitude of induced emf in the conducting loop is 0.99 mV.

Explanation:

Rate of increase in magnetic field per unit time = 0.090 T/s

Area of the conducting loop = 110 cm^2 = 0.0110 m^2

Electromagnetic induction is the production of an emf or voltage in a coil of wire due to a changing magnetic field through the coil.

Induced e.m.f is given as:

EMF = (-N*change in magnetic field/time)*Area

EMF = rate of change of magnetic field per unit time * Area

EMF = 0.090 * 0.0110

EMF = 0.00099 V

EMF = 0.99 mV

5 0
3 years ago
0.16 mol of argon gas is admitted to an evacuated 70 cm^3 container at 30°C. The gas then undergoes an isothermal expansion to a
Semmy [17]

Answer:

The final pressure of the gas is 9.94 atm.

Explanation:

Given that,

Weight of argon = 0.16 mol

Initial volume = 70 cm³

Angle = 30°C

Final volume = 400 cm³

We need to calculate the initial pressure of gas

Using equation of ideal gas

PV=nRT

P_{i}=\dfrac{nRT}{V}

Where, P = pressure

R = gas constant

T = temperature

Put the value in the equation

P_{i}=\dfrac{0.16\times8.314\times(30+273)}{70\times10^{-6}}

P_{i}=5.75\times10^{6}\ Pa

P_{i}=56.827\ atm

We need to calculate the final temperature

Using relation pressure and volume

P_{2}=\dfrac{P_{1}V_{1}}{V_{2}}

P_{2}=\dfrac{56.827\times70}{400}

P_{2}=9.94\ atm

Hence, The final pressure of the gas is 9.94 atm.

3 0
3 years ago
You are trying to find out how high you have to pitch a water balloon in order for it to burst when it hits the ground. You disc
FrozenT [24]

Answer:

The balloon hit the ground with velocity -15.34 m/s

Explanation:

<em>Lets explain how to solve the problem</em>

You found that the best height to pitch a water balloon in order for it to

burst when it hits the ground is 12 meters.

We consider that the 12 meters is the maximum height, so the velocity

at this height is zero.

To find the velocity when the balloon hits the ground lets use the rule

<em>v² = u² + 2gh</em>, where v is the final velocity, u is the initial velocity, g is

the acceleration of gravity and h is the height.

u = 0 , h = 12 m , g = 9.8 m/s²

<em>Substitute these values in the equation above</em>

v² = 0 + 2(9.8)(12)

v² = 235.2

<em>Take square root for both sides</em>

v = ± \sqrt{235.2}

The velocity is downward, then it's a negative value

Then v = -15.34 m/s

<em>The balloon hit the ground with velocity -15.34 m/s</em>

6 0
3 years ago
list 8 countries which have launched satellites into space and also list the name of the satellite which was launched by the cou
Alja [10]
Launch-capable countries
Order Country Satellite(s)
1 Soviet Union Sputnik 1
2 United States Explorer 1
3 France Astérix
4 Japan Ohsumi
10 more rows
7 0
3 years ago
Two spheres have identical charges and are 75 cm apart. the force between them is +0.30 n. what is the magnitude of the charge o
ella [17]

Answer:

4.33\cdot 10^{-6}C, charges are both positive or both negative

Explanation:

The electrostatic force between the two spheres is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the charges on the two spheres

r is the distance between the centres of the two spheres

In this problem, we have

F=+0.30 N is the force

r=75 cm=0.75 m is the distance between the spheres

q_1 =q_2 =q because the two spheres have identical charge

Solving the formula for q, we find

q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(+0.30 N)(0.75 m)^2}{9\cdot 10^9}}=4.33\cdot 10^{-6}C

And the two charges have the same sign (so, both positive or both negative), since the sign of the force is positive (+0.30 N), so it is a repulsive force.

5 0
3 years ago
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