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Basile [38]
3 years ago
6

A heavy box weighing 1000 newtons sits on the floor. You lift upward on the box with a force of 450 newtons, but the box does no

t move. What is the normal force on the box while you are lifting?
Physics
2 answers:
Svetllana [295]3 years ago
5 0
The answer for sure is 550

Ex: 1000 is the box
450 is your force
1000-450=550
Anvisha [2.4K]3 years ago
3 0

Answer:

i think its 550

Explanation:

1000-450 is 550

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The pressure of a sample of gas is measured as 49 torrent. Convert this to atmosphere ​
spin [16.1K]

Answer:

P = 0.0644 atm

Explanation:

Given that,

The pressure of a sample of gas is measured as 49 torr.

We need to convert this temperature to atmosphere.

The relation between torr and atmosphere is as follow :

1 atm = 760 torr

1 torr = (1/760) atm

49 torr = (49/760) atm

= 0.0644 atm

Hence, the presssure of the sample of gas is equal to 0.0644 atm.

5 0
2 years ago
20 points and brainiest
DIA [1.3K]

Answer:

electrical energy sometimes.

Explanation:

125.0m

300 degree Fahrenheit

7 0
3 years ago
You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height (h) from the starting
Eddi Din [679]

Answer:

h = 50.49 m

Explanation:

Data provided:

Speed of skier, u = 2.0 m/s

Maximum safe speed of the skier, v = 30.0 m/s

Mass of the skier, m = 85.0

Total work = 4000 J

Height from the starting gate = h

Now, from the law of conservation of energy

Total energy at the gate = total energy at the time maximum speed is reached

\frac{1}{2}mu^2+mgh=4000J+\frac{1}{2}mv^2

where, g is the acceleration due to the gravity

on substituting the values, we get

\frac{1}{2}\times85\times2.0^2+85\times9.81\times h=4000J+\frac{1}{2}\times85\times30^2

or

170 + 833.85 × h = 4000 + 38250

or

h = 50.49 m

7 0
3 years ago
ball is dropped from rest and hits the ground 1.1 seconds later. What height was the ball released from? ​
Aleonysh [2.5K]
Height = 1/2 * 9.8 * (1.1^2)
5 0
3 years ago
Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad
love history [14]

Answer:

a) The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}, b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}

Where:

a_{r} - Magnitude of the radial acceleration, measured in meters per square second.

a_{t} - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

\| \vec a \| = a_{t}

\| \vec a \| = r \cdot \alpha

Where:

\alpha - Angular acceleration, measured in radians per square second.

r - Radius of rotation (Radius of a tire), measured in meters.

Given that \alpha = 15\,\frac{rad}{s^{2}} and r = 0.31\,m. The linear acceleration experimented by the car is:

\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)

\| \vec a \| = 4.65\,\frac{m}{s^{2}}

The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}.

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

Where:

\theta - Final angular position, measured in radians.

\theta_{o} - Initial angular position, measured in radians.

\omega_{o} - Initial angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

If \theta_{o} = 0\,rad, \omega_{o} = 0\,\frac{rad}{s}, \alpha = 15\,\frac{rad}{s^{2}}, the final angular position is:

\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}

\theta = 46.875\,rad

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

\theta = 7.46\,rev

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0
3 years ago
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