Ask question

Ask question

All categories

2 years ago

5

A 0.5 kg cheeseburger is lobbed at a particularly unhappy customer with a force of 10 N. While it is being thrown, what is the a

cceleration experienced by the cheeseburger? Answer in units of m/s2

1 answer:

Svetllana [295]2 years ago

4 0

Lol i think someone would be fired from their job if they threw food

You might be interested in

A 4.00-kg box sits atop a 10.0-kg box on a horizontal table. The coefficient of kinetic friction between the two boxes and betwe

natta225 [31]

First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.

Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.

4 0

2 years ago

When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ± 17.8 ∘ from the

bazaltina [42]

Look on this website http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html

4 0

2 years ago

If period of the pendulum in preceding sample problem were 24s how tall would the tower be ?

frutty [35]

Answer:

So length of pendulum is 143.129 m

Explanation:

We have given period of simple pendulum is 2 sec

We have to find the length of simple pendulum

Let the length of pendulum is l

Acceleration due to gravity $g=9.8m/sec^2$ is

Time period is given by $T=2\pi \sqrt{\frac{l}{g}}$

So $24=2\times 3.14\times \sqrt{\frac{l}{9.8}}$

$\sqrt{\frac{l}{9.8}}=3.821$

Squaring both side

${\frac{l}{9.8}}=14.60$

l =143.129 m

So length of pendulum is 143.129 m

8 0

2 years ago

Read 2 more answers

A certain humidifier operates by raising water to the boiling point and then evaporating it. Every minute 30 g of water at 20◦ C

Sveta_85 [38]

Answer:

The value of total energy needed per minute for the humidifier = 77.78 KJ

Explanation:

Total energy per minute the humidifier required = Energy required to heat water to boiling point) + Energy required to convert liquid water into vapor at the boiling point) ----- (1)

Specific heat of water = 4190 $\frac{J}{kg k}$

The heat of vaporization is = 2256 $\frac{KJ}{kg}$

Mass = 0.030 kg

Energy needed to heat water to boiling point = $m c ( T_{2} - T_{1} )$

Energy needed to heat water to boiling point = 0.030 × 4.19 × (100 - 20)

Energy ( $E_{1}$ ) = 10.08 KJ

Energy needed to convert liquid water into vapor at the boiling point

$E_{2}$ = 0.030 × 2256 = 67.68 KJ

Thus the total energy needed E = $E_{1}$ + $E_{2}$

E = 10.08 + 67.68

E = 77.78 KJ

This is the value of total energy needed per minute for the humidifier.

9 0

2 years ago

A shopper pushes a grocery cart 41.9 m on level ground, against a 44.5 N frictional force. The cart has a mass of 16.3 kg. He pu

Rashid [163]

Answer:

Fp = 26.59[N]

Explanation:

This problem can be solved using the principle of work and energy conservation, i.e. the final kinetic energy of a body will be equal to the sum of the forces that do work on the body plus the initial kinetic energy.

We need to identify the initial data:

d = distance = 41.9[m]

Ff = friction force = 44.5 [N]

m = mass = 16.3 [kg]

v1 = 1.9 [m/s]

v2 = 12.6 [m/s]

The kinetic energy at the beginning can be calculated as follows:

$E_{k1}= \frac{1}{2}*m*v_{1}^2 \\E_{k1}= \frac{1}{2}*16.3*(1.9)_{1}^2\\E_{k1}= 29.42[J]$

And the final kinetic energy.

$E_{k2}= \frac{1}{2}*m*v_{2}^2 \\E_{k2}= \frac{1}{2}*16.3*(12.6)^2\\E_{k2}= 1294[J]$

The work is performed by two forces, the friction force and the pushing force, it is important to clarify that these forces are opposite in direction.

The weight of the cart also performs a work in the direction of movement since the plane is tilted down, this component of the weight of the cart must be parallel to the surface of the inclined plane.

$W_{1-2}=-(44.5*41.9)+(16.3*9.81*sin(17.5)*41.9)+(F_{p}*41.9) \\therefore:\\E_{k1}+W_{1-2}=E_{k2}\\29.42+150.16+(F_{p}*41.9)=1294\\F_{p}=1114.42/41.9\\F_{p}=26.59[N]$

5 0

2 years ago