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statuscvo [17]
3 years ago
10

You want to build a snowman, so you accelerate a 2kg snowball across your yard at a rate of 0.5m/s2. Calculate the amount of for

ce you applied to your friend.

Physics
1 answer:
tamaranim1 [39]3 years ago
3 0

Answer:

4

Units:

Newtons

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An object will not lose its blank as it moves away from earths surface, but it will lose blank
ladessa [460]
Mass, weight. Mass is independent of the object’s location, whereas weight is not.
3 0
3 years ago
A long solenoid has a radius of 4.0 cm and has 800 turns/m. If the current in the solenoid is increasing at the rate of 3.0 A/s,
kolbaska11 [484]

Answer:

Explanation:

Given that,

Radius of solenoid R = 4cm = 0.04m

Turn per length is N/l = 800 turns/m

The rate at which current is increasing di/dt = 3 A/s

Induced electric field?

At r = 2.2cm=0.022m

µo = 4π × 10^-7 Wb/A•m

The magnetic field inside a solenoid is give as

B = µo•N•I

The value of electric field (E) can

only be a function of the distance r from the solenoid’s axis and it give as,

From gauss law

∮E•dA =qenc/εo

We can find the tangential component of the electric field from Faraday’s law

∮E•dl = −dΦB/dt

We choose the path to be a circle of radius r centered on the cylinder axis. Because all the requested radii are inside the solenoid, the flux-area is the entire πr² area within the loop.

E∮dl = −d/dt •(πr²B)

2πrE = −πr²dB/dt

2πrE = −πr² d/dt(µo•N•I)

2πrE = −πr² × µo•N•dI/dt

Divide both sides by 2πr

E =- ½ r•µo•N•dI/dt

Now, substituting the given data

E = -½ × 0.022 × 4π ×10^-7 × 800 × 3

E = —3.32 × 10^-5 V/m

E = —33.2 µV/m

The magnitude of the electric field at a point 2.2 cm from the solenoid axis is 33.2 µV/m

where the negative sign denotes counter-clockwise electric field when looking along the direction of the solenoid’s magnetic field.

3 0
3 years ago
Read 2 more answers
a quarterback throws a football with angle of elevation 40degrees and speed 60ft/s. Find the horizontal and vertical components
pshichka [43]
<span>
the horizontal velocity would be equal to
Vh = sin (40) /60
= 0.74 * 60
= 44
</span>
the Vertical velocity would be equal to
<span>Vv = cos(40) * 60
=40</span>
4 0
3 years ago
The block accelerates down the 30 degree incline at 3m/s2. What is the coefficient of friction of these surfaces
dolphi86 [110]
Fnet = Fg sin 30 - Ff
ma = mg sin 30 - mew Fg cos 30
ma = mg sin 30 - mew mg cos 30
a = g sin 30 - mew gcos30
a - g sin 30 = - mew g cos 30
mew = -(a - g sin30)/(g cos 30)
mew = -(3m/s2 - 9.81sin30)/(9.81 cos 30)
mew = 0.22
8 0
3 years ago
The specific heat of soil is 0.20 kcal/kg*C and the specific heat of water is 1.00 kcal/kg*C. This means that if 1 kg of soil an
stiv31 [10]

Answer: The soil will be 4\°C warmer than the water.

Explanation:

The heat (thermal energy) absorbed can be found using the following equation:

Q=m.C.\Delta T

Where:

Q is the heat  

m is the mass of the element

C is the specific heat capacity of the material.

\Delta T is the variation in temperature

<u>In the case of soil we have:</u>

Q_{soil}=m_{soil}.C_{soil}.\Delta T_{soil} (1)

Where:

Q_{soil}=1 kcal

m_{soil}=1 kg

C_{soil}=0.2 kcal/kg \°C

\Delta T_{soil}

<u>In the case of water we have:</u>

Q_{water}=m_{water}.C_{water}.\Delta T_{water} (2)

Where:

Q_{water}=1 kcal

m_{water}=1 kg

C_{water}=1 kcal/kg \°C

\Delta T_{water}

Isolating \Delta T from both equations:

\Delta T_{soil}=\frac{Q_{soil}}{m_{soil}.C_{soil}} (3)

\Delta T_{soil}=\frac{1 kcal}{1 kg(0.2 kcal/kg \°C)}

\Delta T_{soil}=5\°C (4)

\Delta T_{water}=\frac{Q_{water}}{m_{water}.C_{water}} (5)

\Delta T_{water}=\frac{1 kcal}{1 kg(1 kcal/kg \°C)}

\Delta T_{water}=1\°C (6)

Comparing (4) and (6) we can find the soil will be 4\°C warmer than the water.

8 0
3 years ago
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